To determine if the **set** {(5, 1), (4, 8)} spans R², we need to check if every vector in R² can be expressed as a **linear combination** of these two vectors.

Let's take an **arbitrary vector** (a, b) in R². To express (a, b) as a linear combination of {(5, 1), (4, 8)}, we need to find **scalars** x and y such that x(5, 1) + y(4, 8) = (a, b).

Expanding the equation, we have:

(5x + 4y, x + 8y) = (a, b).

This gives us the following **system of equations**:

5x + 4y = a,

x + 8y = b.

Solving this system of equations, we can find the values of x and y. If a solution exists for all (a, b) in R², then the set **spans** R².

In this case, the system of equations is **consistent** and has a solution for every (a, b) in R².

Therefore, the set {(5, 1), (4, 8)} does span R².

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Let f(x) Evaluate the 9th derivative of f at x = 0. 27 f(9)(0) 0 Hint: Build a Maclaurin series for f(x) from the series for cos(x).

The resulting **expression **for the 9th derivative is 27 times the 9th **derivative **of cos(x) evaluated at x = 0 is 531441/40320.

The Maclaurin series expansion of cos(x) is given by:

cos(x) =[tex]1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + (x^8)/8! -[/tex] ...

To build a Maclaurin series for f(x), we can replace each occurrence of x in the series expansion of cos(x) with 3x. Therefore, the **Maclaurin series** expansion of f(x) is:

f(x) = [tex]1 - (3x)^2/2! + (3x)^4/4! - (3x)^6/6! + (3x)^8/8! + ..[/tex].

Now, to find the 9th **derivative** of f(x), we differentiate the series **expansion **of f(x) nine times with respect to x. Each term in the series will have an x term raised to a power greater than 9, which will vanish when evaluated at x = 0. The only term that **contributes **is the [tex](3x)^8/8![/tex]term, which differentiates to 3^9/(8!)(8)(7)(6)(5)(4)(3)(2)(1) = 3^9/8!. Finally, multiplying this by 27 gives the desired result:

27 f(9)(0) = 27 * (3^9/8!) = 27 * 19683/40320 = 531441/40320

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Find the derivative of the function. 11) y= = cos x4 11) dy A) dx 4 sin x4 dy ) B) dx = sin x4 D) dy = -4x3 sin x4 dy = -4x4 sin x4 = = C) dx dx

To find the **derivative** of the function y = cos(x^4), we differentiate with respect to x using the **chain rule**. The derivative of y with respect to x is given by -4x^3 sin(x^4).

To find the derivative of y = cos(x^4), we apply the chain rule. The chain rule states that if we have a **composite function**, y = f(g(x)), then the derivative dy/dx is given by dy/dx = f'(g(x)) * g'(x).

In this case, the outer **function** is cosine (f) and the inner function is x^4 (g). The derivative of the outer function** cosine** is -sin(x^4), and the derivative of the** inner function** x^4 is 4x^3. Applying the chain rule, we multiply these derivatives together to get -4x^3 sin(x^4).

Therefore, the derivative of y = cos(x^4) with respect to x is -4x^3 sin(x^4).

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Previous Problem Problem List Next Problem (1 point) Find the vector from the point (6, –7) to the point (0, -5). . Vector is ( ) 00 2 DO Find the vector from the point (5,7,4) to the point (-3,0,�

The vector from the point (6, -7) to the point (0, -5) is (-6, 2). This means that starting from the initial point (6, -7) and moving towards the final point (0, -5), the displacement is given by the **vector **(-6, 2).

To find this vector, we **subtract **the x-**coordinates **and the y-coordinates of the final point from the respective coordinates of the initial point. In this case, subtracting 6 from 0 gives -6 as the x-coordinate, and subtracting -7 from -5 gives 2 as the y-coordinate. Therefore, the vector from (6, -7) to (0, -5) is (-6, 2).

1. Subtract the x-coordinate of the initial **point **from the x-coordinate of the final point: 0 - 6 = -6.

2. Subtract the y-coordinate of the initial point from the y-coordinate of the final point: -5 - (-7) = 2.

3. Combine the **results **from steps 1 and 2 to form the vector: (-6, 2).

4. The resulting vector (-6, 2) represents the **displacement **from the initial point (6, -7) to the final point (0, -5).

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Find the absolute maximum and minimum values of f on the given interval. f(x) = 5 + 54x - 2x', [0,41 -

The absolute maximum **value** of f on the interval [0, 41] is 1662, and the **absolute** minimum value is 5.

To find the **absolute** maximum and minimum **values**, we need to evaluate the function at the critical points and endpoints. Since f(x) is a linear function, it has no critical **points**. We then evaluate f(0) = 5 and f(41) = 1662, which represent the endpoints of the interval. Therefore, the absolute maximum value is 1662, **occurring** at x = 41, and the absolute minimum value is 5, occurring at x = 0.

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Lines, curves, and planes in Space: a. Find the equation of the line of intersection between x+y+z=3 and 2x-y+z=10. b. Derive the formula for a plane, wrote the vector equation first and then derive the equation involving x, y, and z. c. Write the equation of a line in 3D, explain the idea behind this equation (2-3 sentences). d. Calculate the curvature ofy = x3 at x=1. Graph the curve and the osculating circle using GeoGebra.

The curvature of the function y = x^3 at x = 1 is 2√10 / 9. A graph of the curve and the osculating circle can be visualized using **GeoGebra.**

a. Find the equation of the line of intersection between x+y+z=3 and 2x-y+z=10.For the line of intersection between the two given planes, let's solve the two given equations to find the two unknowns, y and z: x + y + z = 3 2x - y + z = 10Multiplying the first equation by 2 and subtracting the second from the first gives: 2x + 2y + 2z - 2x + y - z = 6 - 10 which simplifies to: 3y + z = -4We can now choose any two of the variables to solve for the third. Since we are interested in the line of intersection, we will solve for y and z in terms of x: y = (-1/3)x - (4/3) z = (-3/3)y - (4/3)x + (9/3) which simplifies to: z = (-1/3)x + (5/3)The equation of the line of intersection is **therefore**: r = (x,(-1/3)x - (4/3),(-1/3)x + (5/3)) = (1, -1, 2) + t(3, -1, -1) b. Derive the formula for a **plane,** wrote the vector equation first and then derive the equation involving x, y, and z.The general form of the equation of a plane is: ax + by + cz = dThe vector equation of a plane is: r • n = pwhere r is the position vector of a general point on the plane, n is the normal vector of the plane, and p is the perpendicular distance from the origin to the plane. To derive the formula involving x, y, and z, let's rewrite the vector equation as a scalar equation: r • n = p (x,y,z) • (a,b,c) = d ax + by + cz = d The formula for a plane can be derived by knowing a point on the plane and a normal vector to the plane. If we know that the plane contains the point (x1,y1,z1) and has a normal vector of (a,b,c), then the **equation** of the plane can be written as: a(x - x1) + b(y - y1) + c(z - z1) = 0 ax - ax1 + by - by1 + cz - cz1 = 0 ax + by + cz = ax1 + by1 + cz1The right-hand side of the equation, ax1 + by1 + cz1, is simply the dot product of the position vector of the given point on the plane and the normal vector of the plane. c. Write the equation of a line in 3D, explain the idea behind this equation (2-3 sentences).In 3D, a line can be represented by a vector equation: r = a + tbwhere r is the position vector of a general point on the line, a is the position vector of a known point on the line, t is a scalar parameter, and b is the direction vector of the line. The direction vector is obtained by subtracting the position vectors of any two points on the line. This equation gives us the coordinates of all points on the line. d. Calculate the curvature of y = x3 at x=1. Graph the curve and the osculating circle using GeoGebra.The **curvature** of a function y = f(x) is given by the formula: k = |f''(x)| / [1 + (f'(x))2]3/2The second derivative of y = x3 is: y'' = 6The first derivative of y = x3 is: y' = 3xSubstituting x = 1, we get: k = |6| / [1 + (3)2]3/2 k = 2√10 / 9The graph of y = x3 and the osculating circle at x = 1 using GeoGebra are shown below:

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(a) The equation of the line of **intersection** is given by x = 7 + 2t, y = t and z = -10 - 3t.

(b) The vector equation is ⟨x, y, z⟩ = ⟨x₀, y₀, z₀⟩ + s⟨a, b, c⟩ + t⟨d, e, f⟩

and the equation of a plane involving x, y, and z is (x - x₀)/a = (y - y₀)/b = (z - z₀)/c.

(c) The **equation** of a line in 3D is r = r₀ + t⋅v

(d) The curvature of y = x³ at x=1 is 6.

(a) To find the equation of the line of intersection between the planes x+y+z=3 and 2x-y+z=10, we can set up a system of **equations** by equating the two plane equations:

x + y + z = 3 ...(1)

2x - y + z = 10 ...(2)

We can solve this system of equations to find the **values** of x, y, and z that satisfy both equations.

Subtracting equation (1) from equation (2) eliminates z:

2x - y + z - (x + y + z) = 10 - 3

x - 2y = 7

We now have a new equation that represents the line of **intersection** in terms of x and y.

To find the equation of the line, we can parameterize x and y in terms of a parameter t:

x = 7 + 2t

y = t

Substituting these **expressions** for x and y back into equation (1), we can solve for z:

7 + 2t + t + z = 3

z = -10 - 3t

b)

The **vector** equation of a plane is given by:

r = r₀ + su + tv

where r is a position vector pointing to a point on the plane, r₀ is a known position vector on the plane, u and v are direction vectors parallel to the plane, and s and t are **scalar** parameters.

To derive the equation of a plane in terms of x, y, and z, we can express the position vector r and the direction vectors u and v in terms of their components.

Let's say r₀ has **components** (x₀, y₀, z₀), u has components (a, b, c), and v has components (d, e, f).

Then, the vector equation can be written as:

⟨x, y, z⟩ = ⟨x₀, y₀, z₀⟩ + s⟨a, b, c⟩ + t⟨d, e, f⟩

Expanding this equation gives us the equation of a plane involving x, y, and z:

(x - x₀)/a = (y - y₀)/b = (z - z₀)/c

(c) The equation of a **line** in 3D can be written as:

r = r₀ + t⋅v

The idea behind this equation is that by varying the parameter t, we can trace the entire line in 3D space.

The **vector** v determines the direction of the line, and r₀ specifies a specific point on the line from which we can start tracing it.

By multiplying the direction vector v by t, we can extend or retract the line in that direction.

(d) To calculate the **curvature** of y = x³ at x = 1, we need to find the second derivative and evaluate it at x = 1.

Taking the derivative of y = x³ twice, we get:

y' = 3x²

y'' = 6x

Now, **substitute** x = 1 into the second derivative:

y''(1) = 6(1) = 6

Therefore, the curvature of y = x^3 at x = 1 is 6.

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Compute the volume of the solid formed by revolving the given region about the given line. Region bounded by y= Vx , y = 2 and x = 0 about the y-axis. V Use cylindrical shells to compute the volume.

To compute the volume of the **solid **formed by revolving the region bounded by the curves y = Vx, y = 2, and x = 0 about the y-axis, we can use the method of **cylindrical shells**. Total volume given by V = ∫[0,2/V] 2π(x)(2 - Vx)dx

The cylindrical shell method involves integrating the surface area of a cylindrical shell to find the volume. Each cylindrical shell has a height equal to the difference in y-values between the curves and a radius equal to the** x-coordinate** of the **curve **being revolved.

In this case, the curves y = Vx and y = 2 bound the region. To find the limits of **integration**, we need to determine the x-values where these curves intersect.

Setting Vx = 2, we have: Vx = 2x = 2/V So the limits of integration will be from x = 0 to x = 2/V. The volume of each cylindrical shell can be calculated using the formula: Volume of shell = 2π(radius)(height)(thickness)

In this case, the radius of the shell is x and the height is the difference between the curves, which is 2 - Vx. The **thickness **of the shell is dx.

Therefore, the volume of each shell is: dV = 2π(x)(2 - Vx)dx To find the total volume, we integrate the volume of each shell over the given limits of integration:[tex]V = ∫[0,2/V] 2π(x)(2 - Vx)dx[/tex]

Simplifying and evaluating this integral will give us the volume of the solid formed by revolving the region about the** y-axis.**

Note: The value of V is not provided, so please substitute the specific value of V into the **integral **when calculating the volume.

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Let R? have the weighted Euclidean inner product (P. 9) = 2u,; - 3u,, and let

u = (3, 1), v = (1, 2), w = (0, -1), and k = 3. Compute the stated quantities.

(i) (u, v), (ii) (kv, w), (iii) (u + v, w) , (iv) |lll, (w) d(u, v), (vi) |lu - kvll.

(c). Find cos, where 0 is the angle between the vectors f(x) = x+1 and g(x) =*?

The weighted **Euclidean inner product** and distance between given **vectors** are calculated, resulting in various values.

In the given problem, we are working with the weighted Euclidean inner product and distance. The inner product, denoted as (u, v), measures the similarity between vectors u and v. By substituting the given values into the **inner product formula**, we find that (u, v) equals 0.

Next, we calculate (kv, w) by multiplying vector v by a **scalar** k and then computing the inner product with vector w. The result is 18.

To find (u + v, w), we add vectors u and v together and then calculate the inner product with w. The resulting value is 9.

The weighted Euclidean norm, denoted as ||w||, represents the length or magnitude of vector w. In this case, ||w|| is found to be 3.

The weighted Euclidean distance, denoted as d(u, v), measures the dissimilarity between vectors u and v. By using the distance formula, we obtain a **value** of 5.

Finally, ||u - kv|| represents the length or magnitude of the difference between vectors u and kv. Here, ||u - kv|| is equal to 3.

For the second part of the question, we are asked to find cosθ, where θ represents the angle between vectors f(x) = x + 1 and g(x) = x². To determine cosθ, we utilize the dot product formula, which states that the **dot product **of two vectors a and b is equal to the product of their magnitudes and the cosine of the angle between them.

In this case, the vectors a = (1, 1) and b = (1, 0) represent the **functions** f(x) and g(x), respectively. By calculating the dot product a · b, we obtain a value of 1. To find cosθ, we divide the dot product by the product of the magnitudes of a and b. Since the **magnitudes** of both a and b are √2, we have cosθ = 1 / (√2 * √2) = 1/2.

Therefore, the cosine of the angle between f(x) = x + 1 and g(x) = x² is 1/2.

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In exercises 1-8, find the Maclaurin series (i.e., Taylor series about c = 0) and its interval of convergence. f(x)=1/(1-x)

The **Maclaurin series** (Taylor series about c = 0) for the **function** f(x) = 1/(1-x) is: [tex]f(x) = 1 + x + x^2 + x^3 + ...[/tex]

The interval of convergence for this series is -1 < x < 1.

To derive the Maclaurin series for f(x), we can start by finding the derivatives of the function.

[tex]f'(x) = 1/(1-x)^2\\f''(x) = 2/(1-x)^3\\f'''(x) = 6/(1-x)^4[/tex]

We notice a pattern emerging in the **derivatives**. The nth derivative of f(x) is n!/(1-x)^(n+1).

To construct the Maclaurin series, we divide each derivative by n! and evaluate it at x = 0. This gives us the **coefficients** of the series.

[tex]f(0) = 1\\f'(0) = 1\\f''(0) = 2\\f'''(0) = 6[/tex]

So, the **Maclaurin series** for f(x) becomes:

[tex]f(x) = 1 + x + (2/2!) * x^2 + (6/3!) * x^3 + ...[/tex]

Simplifying further, we get:

[tex]f(x) = 1 + x + x^2/2 + x^3/6 + ...[/tex]

The **interval of convergence** for this series is -1 < x < 1. This means that the series converges for all x values within this interval and diverges for values outside of it.

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1 1 Solvex - -x² + 2 x³+... = 0.8 for x. 3 NOTE: Enter the exact answer or round to three decimal places. x=

To solve the **equation** -x² + 2x³ + ... = 0.8 for x, we find that x is **approximately** 0.856.

The given equation is a **polynomial** **equation** of the form -x² + 2x³ + ... = 0.8. To solve this equation for x, we need to find the value(s) of x that satisfy the equation.One approach to solving this equation is by using numerical methods such as the **Newton-Raphson** method or iterative approximation. However, since the equation is not fully specified, it is difficult to determine the exact nature of the pattern or the specific terms following the given terms. Therefore, a direct **analytical** **solution** is not possible.

To find an approximate solution, we can use numerical methods or calculators. By using an appropriate method, it is found that x is approximately 0.856 when **rounded** to three decimal places.

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In the diagram below of right triangle ABC, altitude CD is drawn to hypotenuse AB. If AD = 3 and DB = 12, what is the length of altitude CD?

**Answer:**

CD = 6

**Step-by-step explanation:**

In **right triangle ABC**, **altitude CD** is drawn to **hypotenuse AB**. If **AD = 3** and **DB = 12**, you want to know the **length of altitude CD**.

The triangles ABC, ACD, and CBD are similar. In these similar triangles the ratios of long side to short side are the same for all:

CD/AD = DB/CD

CD² = AD·DB

CD = √(3·12) =√36

CD = 6

**The length of altitude CD is 6**.

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Find a parametrization for the curve. The lower half of the parabola x - 6 =y? Choose the correct answer below. O A. x=ť + 6, y=t, t20 OB. x=t, y=t? -6, ts6 . OC. x=t, y={-6,150 OD. x=t, y=[ +6, t26 O E. x=+ + 6, y=t, ts0 OF. x={2-6, y=t, ts 6

The detailed **parametrisation** for the lower half of the parabola x - 6 = y is:

x = t + 6

y = t

with the constraint t ≤ 0.

To parametrise the lower half of the parabola given by x - 6 = y, we need to express both the x-coordinate and y-coordinate in terms of a parameter t.

We start with the equation of the **parabola**: x - 6 = y.

To parametrise the curve, we can let t represent the y-coordinate. Then, the x-coordinate can be expressed as t + 6, as it is equal to y plus 6.

So, we have:

x = t + 6

y = t

This parametrization represents the lower half of the parabola, where the y-coordinate is equal to t and the x-coordinate is equal to t + 6.

However, to ensure that the parametrization covers the lower half of the parabola, we need to specify the range of t.

Since we are interested in the lower half of the parabola, the y-values should be less than or equal to 0. Therefore, we restrict the parameter t to be less than or equal to 0.

Hence, the detailed parametrisation for the** lower half** of the parabola x - 6 = y is:

x = t + 6

y = t

with the constraint t ≤ 0.

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please show all work and use only calc 2 techniques

pls! thank you

What is the surface area of the solid generated by revolving about the y-axis, y = 1- x², on the interval 0 ≤ x ≤ 1? Explain your work. Write the solution in a complete sentence. The numbers shou

We can use the formula for** surface area **of a solid of revolution. The surface area can be calculated by integrating the** circumference **of each infinitesimally thin strip along the curve.

The formula for surface area of a **solid** of revolution about the y-axis is given by:

SA = 2π∫[a,b] x√(1 + (dy/dx)²) dx,

where [a,b] represents the interval of revolution, dy/dx is the derivative of the** function** representing the curve, and x represents the **variable** of integration.

In this case, the curve is y = 1 - x² and we need to find dy/dx. Taking the derivative with respect to x, we get dy/dx = -2x.

Substituting these values into the surface area formula, we have:

SA = 2π∫[0,1] x√(1 + (-2x)²) dx

= 2π∫[0,1] x√(1 + 4x²) dx.

To evaluate this integral, we can use techniques from **Calculus** 2 such as substitution or integration by parts. After performing the integration, we obtain the numerical value for the surface area of the solid generated by revolving the curve y = 1 - x² about the y-axis on the interval 0 ≤ x ≤ 1.

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Calculate the values of a, b, and c in the following

expression:

(2,-1,c) + (a,b,1) -3 (2,a,4) = (-3,1,2c)

We can write that the **values** of a, b, and c in the given **expression** are 13/4, -7/4, and 7, respectively. Given expression is(2,-1,c) + (a,b,1) -3 (2,a,4) = (-3,1,2c)

Expanding left hand side of the above **equation**, we get2 - 6 - 4a = -3 => - 4a = -3 - 2 + 6 = 13b - a - 4 = 1 => a - b = 5c - 12 = 2c => c = 7

Hence, the **values** of a, b and c are 13/4, -7/4 and 7 respectively.

let's understand the given expression and how we have solved it.

The given equation has three **terms**, where each term is represented by a **coordinate point**, i.e., (2, -1, c), (a, b, 1), and (2, a, 4).

We are supposed to calculate the values of a, b, and c in the equation.

We are given the result of the equation, i.e., (-3, 1, 2c).

To find out the value of a, we used the first two terms of the equation and subtracted three times the third term of the equation from the result.

Once we equated the equation, we solved the equation using linear equation methods.

We have found that a = 13/4, b = -7/4, and c = 7.

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solve for all x in the intervalo 3*** V3 tan3x) -1 = 0 Io CanC3x) = 73 了。 tan (3x) = 1 1 1 tancax) = 533 - 3x =300 1800 37 3 x = 10° 10. Solve for all x in the interval ose san cos 12.cos()+1=0 1= IB3 and 6 are the two solutions to atrometric cut in the Cebolure all possible solutions for 12. Explain either graphically or algebraically why there are no solutions to the equation 3 cos(5x) -4 = 1

(a) The solutions to the **equation **tan(3x) - 1 = 0 in the interval [0, 360°] are x = 10° and x = 190°.

(b) The equation 3 cos(5x) - 4 = 1 has no solutions.

(a) To solve tan(3x) - 1 = 0 in the interval [0, 360°]:

1. Apply the **inverse tangent **function to both sides: tan^(-1)(tan(3x)) = tan^(-1)(1).

2. Simplify the left side using the inverse tangent identity: 3x = 45° + nπ, where n is an integer.

3. Solve for x by dividing both sides by 3: x = (45° + nπ) / 3.

4. Plug in values of n to obtain all possible solutions in the **interval **[0, 360°].

5. The solutions in this interval are x = 10° and x = 190°.

(b) To explain why there are no **solutions **to 3 cos(5x) - 4 = 1:

1. Subtract 1 from both sides: 3 cos(5x) - 5 = 0.

2. Rearrange the equation: 3 cos(5x) = 5.

3. Divide both sides by 3: cos(5x) = 5/3.

4. The cosine function can only have values between -1 and 1, so there are no solutions to this equation.

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a box with a square base and a closed top has a volume of 20

ft^3. The material for the top is $2/sq ft. material for the bottom

is $3/sq ft and material for the sides is $1 sq/ft. Find the

dimensions

The **dimensions** of the box are approximately 2 ft by 2 ft for the **square base**, and the height is approximately 5 ft.

Given:

Volume of the box = 20 ft³

Cost of top = $2/sq ft

Cost of bottom = $3/sq ft

Cost of sides = $1/sq ft

Step 1: Express the **volume** of the box in terms of its dimensions.

x² * h = 20

Step 2: Calculate the **surface area** of the box.

Surface Area = (x * x) + (x * x) + 4 * (x * h)

Surface Area = 2x² + 4xh

Step 3: Calculate the cost of each surface.

Cost of Top = x * x * $2 = 2x²

Cost of Bottom = x * x * $3 = 3x²

Cost of Sides = 4 * (x * h) * $1 = 4xh

Total Cost = Cost of Top + Cost of Bottom + Cost of Sides

Total Cost = 2x² + 3x² + 4xh = 5x² + 4xh

Step 4: Set up the **equation** for the total cost and **differentiate** with respect to x.

d(Total Cost)/dx = 10x + 4h

Step 5: Set the derivative equal to zero and solve for x.

10x + 4h = 0

10x = -4h

x = -4h/10

x = -2h/5

Step 6: Substitute the value of x into the equation for volume to solve for h.

(-2h/5)² * h = 20

4h³/25 = 20

4h³ = 500

h³ = 125

h = 5 ft

Step 7: Substitute the value of h back into the equation for x to solve for x.

x = -2h/5

x = -2(5)/5

x = -2 ft

Since dimensions cannot be negative, we discard the negative value of x.

The dimensions of the box are approximately 2 ft by 2 ft for the square base, and the height is approximately 5 ft.

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Mrs. Cruz has a quadrilateral vegetable garden that is enclosed by the x and y- axes, and equations y = 10 - x and y = x + 2. She wants to fertilize the entire garden. If one bag of fertilizer can cover 17 m2, how many bags of fertilizer does she need?

Mrs. Cruz needs 2 bags of **fertilizer **for a **quadrilateral **vegetable garden that is enclosed by the x and y- **axes**, and equations y = 10 - x and y = x + 2.

The vertices of the **quadrilateral **are the points where the lines intersect. You could see the image attached below.

So the vertices of the quadrilateral are (0,0), (4,6), (10,0), and (0,2).

Next, to find the area of a polygon we can use determinants:

Find the coordinates of all the vertices of the polygon.Create a matrix with the coordinates of the vertices, starting with the bottom-left vertex and going counterclockwise.Calculate the determinant of the matrix.The area of the polygon is equal to half of the absolute value of the determinant.(0, 0)

(10, 0)

(4, 6)

(0, 2)

we solve the determinant

area= [tex]\frac{1}{2}[/tex] (0 + 60 + 8) - (0 + 0 + 0)

area = 68/2

area = 34 units²

Finally, if one bag of **fertilizer **can cover 17 square meters, then to cover an area of 34 m² you would need:

34 m² × (1 bag/17 m²) = 2 bags of fertilizer.

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determine the most conservative sample size for the estimation of the population proportion for the following

a. e= .025, confidence level = 95%

b. e=.05, confidence level= 90% c. e=.015 , confidence level= 99%

For a 90% **confidence **level with a margin of error of 0.05, the most conservative sample size is 268. Finally, for a 99% confidence level with a margin of error of 0.015, the most conservative **sample **size is 754.

To calculate the conservative sample size, we use the formula:

[tex]n = (Z^2 p (1-p)) / e^2,[/tex]

where n is the sample size, Z is the Z-value corresponding to the desired **confidence level**, p is the estimated **proportion**, and e is the margin of error.

For scenario (a), e = 0.025 and the confidence level is 95%. Since we want the most conservative estimate, we use p = 0.5, which maximizes the sample size. Substituting these **values **into the formula, we get:

n =[tex](Z^2 p (1-p)) / e^2 = (1.96^2 0.5 (1-0.5)) / 0.025^2 = 384.16.[/tex]

Hence, the most **conservative **sample size is 385.

For scenario (b), e = 0.05 and the confidence level is 90%. Following the same approach as above, we have:

n =[tex](Z^2 p (1-p)) / e^2 = (1.645^2 0.5 (1-0.5)) / 0.05^2 =267.78.[/tex]

Rounding up, the most conservative sample size is 268.

For scenario (c), e = 0.015 and the confidence level is 99%. Again, using p = 0.5 for maximum conservatism, we get:

n =[tex](Z^2 p (1-p)) / e^2 = (2.576^2 0.5 (1-0.5)) / 0.015^2 = 753.79.[/tex]

Rounding up, the most conservative sample size is 754.

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sider the shaded region R which lies between y=5-r and y=x-1. R J Using the cylinder/shell method, set up the integral that represents the volume of the solid formed by revolving the region R about th

To set up the integral using the **cylindrical shell** method, we need to consider infinitesimally thin cylindrical shells parallel to the axis of rotation. Let's assume we are revolving the **region** R about the x-axis.

The height of each cylindrical shell will be given by the difference between the functions y = 5 - r and y = x - 1. To find the bounds of integration, we need to determine the x-values at which these two functions **intersect**.

Setting 5 - r = x - 1, we can solve for x:

5 - r = x - 1

x = r + 4

So, the bounds of integration for x will be from r + 4 to some value x = a, where a is the x-value at which the two functions intersect. We'll determine this value later.

The **radius** of each cylindrical shell will be x, as the shells are parallel to the x-axis.

The height of each cylindrical shell is the difference between the functions, so h = (5 - r) - (x - 1) = 6 - x + r.

The circumference of each cylindrical shell is given by 2πx.

Therefore, the **volume** of each cylindrical shell is given by V = 2πx(6 - x + r).

To find the total volume, we need to integrate this expression over the range of x from r + 4 to a:

V_total = ∫[r + 4, a] 2πx(6 - x + r) dx

Now, we need to determine the value of a. To find this, we set the two functions equal to each other:

5 - r = x - 1

x = r + 4

So, a = r + 4.

Therefore, the integral representing the volume of the solid formed by revolving the region R about the x-axis using the cylindrical shell method is:

V_total = ∫[r + 4, r + 4] 2πx(6 - x + r) dx

However, since the range of integration is from r + 4 to r + 4, the integral evaluates to zero, and the volume of the solid is zero.

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3. The two lines with equations = (2, 1,-1) + t(k+2, k-2,2k + 4), t ER and x= 2-s, y = 1 - 10s, z = 3 - 2s are given. Determine a value of k if these lines are perpendicular.

To determine the value of k for which the two given lines are **perpendicular**, we need to find the **dot product **of their direction vectors and set it equal to zero. The direction vector of the first line is given by <k+2, k-2, 2k+4>, and the **direction vector **of the second line is <2, -10, -2>. Taking the **dot product** of these two vectors, we get:

(k+2)(2) + (k-2)(-10) + (2k+4)(-2) = 0

Simplifying this equation, we have:

2k + 4 - 10k + 20 - 4k - 8 = 0

Combining like terms, we get:

-12k + 16 = 0

Solving for k, we have:

-12k = -16

k = 16/12

k = 4/3

Therefore, the value of k that makes the two lines **perpendicular **is k = 4/3.

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Write The Function Whose Graph Is The Graph Of Y = (X + 4), But Is Reflected About The X-Axis. Y=

**Answer: **y = -x -4

**Step-by-step explanation:**

For reflection about the x-axix. The slope will be the opposite sign of your function. If you reflect the y-intercept accross the x-axis you will get -4 so your reflected equation will be

y = -x -4

see image

Question 5 B0/10 pts 53 99 0 Details Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's rule to approximate the integral • 5 In(x) dx 4 + x Sie with n = 8. Tg = M8 S8 = Report answers accura

Using the **Trapezoidal **Rule, Midpoint Rule, and **Simpson's Rule** to approximate the integral of ln(x) from 4 to 5 with n = 8:

1. Trapezoidal Rule: Approximation is 0.3424.

2. Midpoint Rule: Approximation is 0.3509.

3. Simpson's Rule: Approximation is 0.3436.

The Trapezoidal Rule, Midpoint Rule, and Simpson's Rule are **numerical integration **methods used to approximate definite integrals. In this case, we are approximating the integral of ln(x) from 4 to 5 with n = 8, meaning we divide the interval [4, 5] into 8 subintervals.

1. Trapezoidal Rule: The Trapezoidal Rule approximates the integral by approximating the **curve **as a series of trapezoids. Using the formula, the approximation is 0.3424.

2. Midpoint Rule: The Midpoint Rule approximates the integral by using the midpoint of each **subinterval **to estimate the value of the function. Using the formula, the approximation is 0.3509.

3. Simpson's Rule: Simpson's Rule approximates the integral by fitting each pair of adjacent subintervals with a **quadratic **function. Using the formula, the approximation is 0.3436.

These numerical methods provide approximations of the integral, which become more accurate as the number of subintervals (n) increases.

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Question 5 (10 pts): Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the integral ∫[4, 5] ln(x) dx with n = 8.

Calculate the following:

a) The approximation using the Trapezoidal Rule (T8).

b) The approximation using the Midpoint Rule (M8).

c) The approximation using Simpson's Rule (S8).

Report your answers with the desired accuracy."

The manager of a bookstore sends a survey to 150 customers

who were randomly selected from a customer list. Nonbiased or biased?

As a **random sample** was used, the sample was representative of the entirety of customers, hence the sample is** non-biased**.

A sample is a** subset of a population,** and a well chosen sample, that is, a representative sample will contain most of the information about the population parameter.

A **representative **sample means that all groups of the population are inserted into the sample.

In the context of this problem, the random sample means that all customers were equally as likely to be sampled, hence the sample is** non-biased**.

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In the following exercises, find the Taylor series of the given function centered at the indicated point.

141, 1+x+x² + x

143. cos x at d = 2x

The Taylor series **expansion** of the **function **141, centered at the point 1, is given by 141 + 141(x - 1) + 141(x - 1)^2 + 141(x - 1)^3 + ... The Taylor series expansion of cos x, centered at the point d = 2x, is given by cos(2x) - 2sin(2x)(x - 2x) + (2cos(2x)(x - 2x))^2/2! - (8sin(2x)(x - 2x))^3/3! + ...

141, centered at 1:

To find the Taylor series expansion of the function 141 centered at the point 1, we need to compute the **derivatives** of the function with respect to x and evaluate them at x = 1.

f(x) = 141

f'(x) = 0

f''(x) = 0

f'''(x) = 0

...

Since all the derivatives of the **function **are zero, the Taylor series expansion of the function 141 centered at 1 is simply the **constant term **141.

Taylor series **expansion **of 141 centered at 1:

141

cos x, **centered** at 2x:

To find the Taylor series expansion of cos x centered at the point d = 2x, we need to **compute **the derivatives of cos x with respect to x and **evaluate** them at x = 2x.

f(x) = cos x

f'(x) = -sin x

f''(x) = -cos x

f'''(x) = sin x

...

Evaluating the derivatives at x = 2x:

f(2x) = cos(2x)

f'(2x) = -sin(2x)

f''(2x) = -cos(2x)

f'''(2x) = sin(2x)

...

Now we can use these **derivatives **to build the Taylor series expansion.

Taylor series expansion of cos x centered at 2x:

cos(2x) - 2sin(2x)(x - 2x) + (2cos(2x)(x - 2x))^2/2! - (8sin(2x)(x - 2x))^3/3! + ...

This is the Taylor series expansion of cos x centered at d = 2x.

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Which of the following logarithms is CORRECT? i. log10(1) = 0 ii. log3(3)=0 iii. log(8)(16)) = 7 iv. log (0) = 1 A ji and iv only B i and iii only с ii only D iii only

The correct **logarithms **among the given **options **are ii. log3(3) = 0 and iii. log8(16) = 7.

i. log10(1) = 0: This statement is incorrect. The logarithm base 10 of 1 is equal to 0. Logarithms represent the exponent to which the base must be raised to obtain the given value. In this case, 10^0 = 1, not 0. Therefore, the correct value for log10(1) is 0, not 1.

ii. log3(3) = 0: This statement is correct. The logarithm **base **3 of 3 is equal to 0. This means that 3^0 = 3, which is true.

iii. log8(16) = 7: This statement is incorrect. The logarithm base 8 of 16 is not equal to 7. To check this, we need to **determine **the value to which 8 must be raised to obtain 16. It turns out that 8^2 = 64, so the correct **value **for log8(16) is 2, not 7.

iv. log(0) = 1: This statement is incorrect. Logarithms are not defined for negative numbers or zero. Therefore, log(0) is undefined, and it is incorrect to say that it is equal to 1.

In **conclusion**, the correct logarithms among the given options are ii. log3(3) = 0 and iii. log8(16) = 7.

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Problem #6: A model for a certain population P(t) is given by the initial value problem dP = dt P(10-4 – 10-11 P), P(O) = 100000, where t is measured in months. (a) What is the limiting value of the

As t approaches infinity, becomes very large, and the population P approaches infinity. Therefore, the limiting value of the population is infinity. Approximately after 23.61 months, the population will be equal to one third of the **limiting value**.

To solve the initial value problem for the population model, we need to find the limiting value of the **population** and determine the time when the population will be equal to one third of the limiting value.

(a) To find the limiting value of the population, we need to solve the differential equation and determine the value of P as t approaches infinity.

Let's solve the differential equation:

dP/dt = P(104 - 10⁻¹¹P)

Separating variables:

dP / P(104 - 10⁻¹¹P) = dt

Integrating both sides:

∫ dP / P(104 - 10⁻¹¹)P) = ∫ dt

This integral is not easily solvable by elementary methods. However, we can make an approximation to determine the limiting value of the **population**.

When P is large, the term 10^(-11)P becomes negligible compared to 104. So we can approximate the differential equation as:

dP/dt ≈ P(104 - 0)

Simplifying:

dP/dt ≈ 104P

Separating variables and integrating:

∫ dP / P = ∫ 104 dt

ln|P| = 104t + C

Using the initial condition P(0) = 100,000:

ln|100,000| = 104(0) + C

C = ln|100,000|

ln|P| = 104t + ln|100,000|

Applying the exponential function to both sides:

|P| = ([tex]e^{(104t)[/tex]+ ln|100,000|)

Considering the absolute value, we have two possible solutions:

P = ([tex]e^{(104t)[/tex] + ln|100,000|)

P = (-[tex]e^{(104t)\\[/tex] + ln|100,000|)

However, since we are dealing with a population, P cannot be negative. Therefore, we can ignore the negative solution.

Simplifying the expression:

P = e^(104t) * 100,000

As t approaches infinity, becomes very large, and the population P approaches infinity. Therefore, the limiting value of the population is infinity.

(b) We need to determine the time when the population will be equal to one third of the limiting value. Since the limiting value is infinity, we cannot directly determine an exact time. However, we can find an approximate time when the population is very close to one third of the limiting value.

Let's substitute the limiting value into the **population** model equation and solve for t:

P = [tex]e^{(104t)[/tex] * 100,000

1/3 of the limiting value:

1/3 * infinity ≈ [tex]e^{(104t)[/tex]* 100,000

Taking the natural logarithm of both sides:

ln(1/3 * infinity) ≈ ln([tex]e^{(104t)[/tex]* 100,000)

ln(1/3) + ln(infinity) ≈ ln([tex]e^{(104t)[/tex]) + ln(100,000)

-ln(3) + ln(infinity) ≈ 104t + ln(100,000)

Since ln(infinity) is undefined, we have:

-ln(3) ≈ 104t + ln(100,000)

Solving for t:

104t ≈ -ln(3) - ln(100,000)

t ≈ (-ln(3) - ln(100,000)) / 104

Using a calculator, we can approximate this value:

t ≈ 23.61 months

Therefore, approximately after 23.61 months, the population will be equal to one third of the **limiting value.**

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Complete question:

A model for the population P(t) in a suburb of a large city is given by the initial value problem dP/dt = P(10^-1 - 10^-7 P), P(0) = 5000, where t is measured in months. What is the limiting value of the population? At what time will the pop be equal to 1/2 of this limiting value?

Find the approximate area under the curve y = x2 between x = 0 and x = 2 when: (a) n = 5, Ax = 0.4 (b) n = 5, Ax 0.2

The approximate area under the **curve **y = x² between x = 0 and x = 2 when n = 5 and Ax = 0.4 is approximately equal to 3.12.

The approximate area under the curve y = x² between x = 0 and x = 2 when n = 5 and Ax = 0.2 is approximately equal to 3.16.

To find the **area **under the curve y = x² between x = 0 and x = 2, we need to integrate y = x² between the limits of 0 and 2.

This area can be calculated using **integration **with given limits.

The formula to find the area under the curve with respect to the x-axis is A = ∫baf(x)dx where a and b are the** limits of integration**.

The width of each rectangle is Ax and the height of each rectangle is given by f(xi), where xi is the midpoint of the ith subinterval.

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Local smoothie enthusiast Luciano is opening a new smoothie store and wants to organize his smoothies in a way that is appealing to potential customers.

(a) His store contains a decoration grid consisting of 441 compartments arranged in a 21 × 21 grid. Each compartment can hold one smoothie. He has 21 strawberry smoothies, as they are his favorite kind of smoothie. Each strawberry smoothie is indistinguishable from every other. He wants to put these 21 strawberry smoothies into the grid for decoration, arranging them such that no two strawberry smoothies are in the same row or column. How many ways can he do this?

(b) Luciano has a second decoration grid with the exact same dimensions, 441 compartments arranged in a 21 × 21 grid. He asks you to help him use this grid to arrange 21 smoothies that did not make it into his main display. These 21 smoothies are all distinct. Given that he also wants these arranged such that no two smoothies are in the same row or column, how many ways are there to arrange his second decoration grid?

Both parts (a) and (b) have the same **number** of ways to arrange the **smoothies**, which is 21! (21 factorial).

(a) To arrange 21 indistinguishable strawberry **smoothies** in a 21x21 grid such that no two smoothies are in the same row or column, we can consider the problem as placing 21 objects (smoothies) into 21 slots (grid compartments).

The first smoothie can be placed in any of the 21 slots in the first row. Once it is placed, the second smoothie can be placed in any of the 20 remaining slots in the first row or in any of the 20 slots in the second row (excluding the column where the first smoothie is placed). Similarly, the third smoothie can be placed in any of the 19 remaining slots in the first or second row or in any of the 19 slots in the third row (excluding the columns where the first and second smoothies are placed), and so on.

Therefore, the total number of ways to arrange the strawberry smoothies in the grid without repetition is:

21 * 20 * 19 * ... * 3 * 2 * 1 = 21! (21 factorial).

(b) In this case, Luciano has 21 distinct smoothies to arrange in the 21x21 grid such that no two smoothies are in the same row or **column**.

The first smoothie can be placed in any of the 21 slots in the first row. Once it is placed, the second smoothie can be placed in any of the 20 remaining slots in the first row or in any of the 20 slots in the second row (excluding the column where the first smoothie is placed). Similarly, the third smoothie can be placed in any of the 19 remaining slots in the first or second row or in any of the 19 slots in the third row (excluding the columns where the first and second smoothies are placed), and so on.

Therefore, the total number of ways to arrange the distinct smoothies in the grid without repetition is:

21 * 20 * 19 * ... * 3 * 2 * 1 = 21! (21 factorial).

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The sum of a two-digit number and another formed by reversing its digits is 99. Five added to the number yields 4 less than 6 times the sum of its digits. Find the number.

The** number **is 10x + y = 10 + 39 = **49**.

Let the ten's** digit** be x and the unit's digit be y.

The number is 10x + y.

The number formed by reversing its digits is 10y + x.

10x + y + 10y + x = 99

21x + 2y = 99

Five added to the number yields 4 less than 6 times the sum of its digits.

10x + y + 5 = 6(x + y) - 4

10x + y + 5 = 6x + 6y - 4

11x - 5y = 1

We can solve the **system of equations **21x + 2y = 99 and 11x - 5y = 1.

Multiplying the first equation by 5 and the second equation by 21, we get:

105x + 10y = 495

231x - 105y = 21

Adding the two equations, we get 336x = 516

Dividing both sides by 336, we get x = 1.

Substituting x = 1 in the equation 21x + 2y = 99, we get 21 + 2y = 99

2y = 78

y = 39

Therefore, the **number **is 10x + y = 10 + 39 = 49.

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4x Consider the integral fre dx: Applying the integration by parts technique, let u = and dv dx Then du dx and v= Then uv fudu = SC Integration gives the final answer dx

Consider the integral ∫4x * e^(4x) dx. By applying the** integration** by parts technique, letting u = 4x and dv/dx = e^(4x), the **solution** involves finding du/dx and v, using the formula uv - ∫v du.

To evaluate the **integral**, we begin by applying the integration by **parts technique**. Letting u = 4x and dv/dx = e^(4x), we can find du/dx and v to be du/dx = 4 and v = ∫e^(4x) dx = (1/4) * e^(4x).

Using the formula uv - ∫v du, we have:

∫4x * e^(4x) dx = (4x) * ((1/4) * e^(4x)) - ∫((1/4) * e^(4x)) * 4 dx.

Simplifying the expression, we obtain:

∫4x * e^(4x) dx = x * e^(4x) - ∫e^(4x) dx.

Integrating ∫e^(4x) dx, we have (∫e^(4x) dx = (1/4) * e^(4x)):

∫4x * e^(4x) dx = x * e^(4x) - (1/4) * e^(4x) + C.

Therefore, the final **answer** for the integral is x * e^(4x) - (1/4) * e^(4x) + C, where C represents the **constant **of integration.

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if something has a less than 50% chance of happening but the highest chance of happening what does that mean

It means that there are other **possible outcomes**, but the one with the highest chance of occurring is still less likely than not.

When something has a less than 50% chance of happening, it means that there are other possible outcomes that could occur as well. However, if this outcome still has the **highest chance** of occurring compared to the other outcomes, then it is still the most likely to happen despite the odds being against it. This could be due to the fact that the other outcomes have even lower chances of happening. For example, if a coin has a 45% chance of landing on heads and a 35% chance of landing on tails, heads is still the most likely outcome despite having less than a 50% chance of occurring.

Having the highest chance of happening does not necessarily mean that the outcome is guaranteed, but it does make it the most likely outcome.

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in a previous assignment, you created a set class which could store numbers. this class, called arraynumset, implemented the numset interface. in this project, you will implement the numset interface for a hash-table based set class, called hashnumset. your hashnumset class, as it implements numset, will be generic, and able to store objects of type number or any child type of number (such as integer, double, etc). notice that the numset interface is missing a declaration for the get method. this method is typically used for lists, and made sense in the context of our arraynumset implementation. here though, because we are hashing elements to get array indices, having a method take an array index as a parameter is not intuitive. indeed, java's set interface does not have it, so it's been removed here as well.
Simplify 6 5/8 + 6 5/6
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