write an equation of the line passing through the point $\left(5,\ -3\right)$ that is parallel to the line $y=x 2$ .

The last entry in the synthetic division table is not zero, 3i is not a zero of the polynomial function g(x) = x³ - 4x² + 9x - 36.

To determine if 3i is a zero of the polynomial function g(x) = x³ - 4x² + 9x - 36, we can use synthetic division.

First, we set up the synthetic division table:

   3i | 1   -4   9   -36

Performing the synthetic division:

      | (1)  (-4)   (9)   (-36)

   3i |        3i    9i²   27i

      | (1)  (-4 + 3i)   (9 + 9i²)   (-36 + 27i)

Simplifying the last row, we have:

      | (1)  (-4 + 3i)   (9 - 9)   (-36 + 27i)

      | (1)  (-4 + 3i)   (0)   (-36 + 27i)

Therefore, the last entry in the synthetic division table is not zero, 3i is not a zero of the polynomial function g(x) = x³ - 4x² + 9x - 36.

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Answer:

NLE Choppa noooooooooooooooooo

Step-by-step explanation:

THE BEST RAPPPERRR!!!!!!!!!

Answer:

He wasnt in jail? I dont think cuz he was makin new music

Step-by-step explanation:

Step-by-step explanation:

V≈75.4

A≈100.53

Answer:

It has been answered for you

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Answer:

This example is not a random example due to the fact that it is only questioning the eighth-graders. Random sample means that the sample is chosen simply randomly and everyone has an equal opportunity to be apart of the sampling.

Step-by-step explanation:

Answer:

V≈7853.98

Step-by-step explanation:

V=πr2h

r=10

h=25

Solution

V=πr2h=π·102·25≈7853.98163

Answer:

2. there is no E it would be AB=CD

Answer:

1

EXPLAINATION:

y = mx +b

b = y intercept

Answer:

3

Step-by-step explanation:

12/4=3

a The net income is $3505000

b The number of common shares outstanding is 399,000

c Earnings per share is $8.79

d Book value of common stock is $46.62

e. Price-earnings ratio (P/E ratio) is 6.14

a. Net income available to common stockholders:

Net income = $3,595,000

Dividends to preferred stockholders = (Number of preferred shares * Par value * Dividend rate) = (10,000 * $100 * 0.09) = $90,000

Net income available to common stockholders = Net income - Dividends to preferred stockholders

= $3,595,000 - $90,000

= $3,505,000

b. Weighted average number of common shares outstanding:

Number of common shares outstanding = 399,000

c. Earnings per share: EPS = Net income available to common stockholders / Weighted average number of common shares outstanding

= $3,505,000 / 399,000

≈ $8.79

d. Book value of common stock:

Total stockholders' equity = $19,600,000

Preferred stock equity = Number of preferred shares * Par value = 10,000 * $100 = $1,000,000

Common stock equity = Total stockholders' equity - Preferred stock equity

= $19,600,000 - $1,000,000

= $18,600,000

Book value of common stock = Common stock equity / Number of common shares outstanding

= $18,600,000 / 399,000

≈ $46.62

e. Price-earnings ratio (P/E ratio):

Price-earnings ratio = Price per share / Earnings per share

= $54 / $8.79

≈ 6.14

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$7 per hour plus $9 off

$7x5=$35-$9 discount=$26

$26+$7=$33

6hours=$42-$9=$33

Joe could rent the boat for 6 hours

Answer:

We meet again

Step-by-step explanation:

When a rectangle is dilated by a scale factor k, the length and width of the rectangle are both multiplied by k. In this case, the length of the rectangle is being dilated by a scale factor of 5. So the length of the rectangle after the dilation will be 20 * 5 = **100 meters**.

Equation system that corresponds to the graph:

1. y ≤ 3x 2. y > –2x – 1

To find the system of equations that corresponds to the provided graph, we must first analyze it and locate the regions that fulfil the specified requirements.

1. Begin by locating the darkened region underneath the line y 3x. This line has a slope of 3 and intersects the origin (0,0). Shade the area beneath the line.

2. After that, locate the darkened region above the line y > -2x - 1. The slope of this line is -2, while the y-intercept is -1. The area above the line should be shaded.

3. The solution space that meets both requirements is represented by the overlapping shaded region between the two lines. The common area is located below y 3x and above y > -2x - 1.

4. The equation system that corresponds to this common area is: - y 3x - y > -2x - 1

The space for the addition in the open area next to the present building is defined by these two formulae.

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Answer:

28-8-4x=0

x=5

5 days

Step-by-step explanation:

28-8=20 problems left

then 4 each day

Answer:

The angle is 58 degrees

Step-by-step explanation:

Given

See attachment for kite

Required

The angle at the tail end

Represent this angle with x.

From the attached kite, we have:

1 angle = 122

2 angles = right-angled

So, we have:

[tex]x + 122 +90+90 = 360[/tex] --- sum of angles in a kite

[tex]x + 302 = 360[/tex]

Solve for x

[tex]x =- 302 + 360[/tex]

[tex]x =58^\circ[/tex]

Answer:

the slope is 3

the y intercept is (0,-7)

Step-by-step explanation:

3x-y=7 solve for y

add y to both sides

3x=7+y

subtract 7 from both sides

y=3x-7

y=mx+b

m=slope

b=y-intercept

3=slope

-7=y-intercept

Hope that helps :)

Answer:

it's arm will regenerate and grow back

plz mark me as brainliest

Answer:

B. The arm will regenterate

Answer: should be 7819.20

explanation:

7200 * .043 * 2 = 619.20

7200 + 619.20 = 7819.20

(1) To find the value of B that makes f_(XY)(x,y) a valid joint density function, we need to ensure that the total probability over the entire domain is equal to 1. In this case, the domain is |x|<1 and |y|<1.

The integral of f_(XY)(x,y) over the given domain should be equal to 1:

∫∫ f_(XY)(x,y) dx dy = 1

∫∫ B(1+xy) dx dy = 1

To solve this integral, we integrate with respect to x first and then with respect to y:

∫(∫ B(1+xy) dx) dy

∫[Bx + B(xy^2)/2] dy, integrating with respect to x

Bxy + B(xy^2)/2 + C, integrating with respect to y

Now, evaluate the integral over the given domain:

∫[-1,1] [Bxy + B(xy^2)/2 + C] dy

[Bxy^2/2 + B(xy^3)/6 + Cy] evaluated from -1 to 1

[B/2 + B/6 + C] - [-B/2 - B/6 - C]

(B/2 + B/6 + C) - (-B/2 - B/6 - C)

2B/3 = 1

Solving for B:

B = 3/2

Therefore, the value of B that makes f_(XY)(x,y) a valid joint density function is B = 3/2.

(2) To determine if X and Y are uncorrelated, we need to calculate the covariance between X and Y. If the covariance is zero, then X and Y are uncorrelated.

Cov(X, Y) = E[XY] - E[X]E[Y]

To calculate E[XY], we need to find the joint expectation:

E[XY] = ∫∫ xy f_(XY)(x,y) dx dy

E[XY] = ∫∫ xy (3/2)(1+xy) dx dy

Integrating over the domain |x|<1 and |y|<1, we can calculate E[XY].

Similarly, we need to calculate E[X] and E[Y] to determine Cov(X, Y).

If Cov(X, Y) is found to be zero, then X and Y are uncorrelated.

(3) To prove or disprove independence between X and Y, we need to check if the joint probability density function (pdf) can be factorized into the product of the marginal pdfs of X and Y.

If f_(XY)(x,y) = f_X(x)f_Y(y), then X and Y are independent.

To determine if this factorization holds, we need to compare the joint pdf f_(XY)(x,y) with the product of the marginal pdfs f_X(x) and f_Y(y). If they are equal, then X and Y are independent. Otherwise, they are dependent.

(4) To prove or disprove the independence between X^2 and Y^2, we follow a similar approach as in (3). We compare the joint pdf of X^2 and Y^2 with the product of their marginal pdfs. If they are equal, X^2 and Y^2 are independent. Otherwise, they are dependent.

By examining the factorization of the joint pdfs and comparing them with the product of the marginal pdfs, we can determine the independence relationships between the variables X, Y, X^2, and Y^2.

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Answer:

16 cans

Step-by-step explanation:

Hey,

I have to admit, this problem is pretty complicated. But I got you :).

To begin, we have to represent the small and large boxes with two separate variables. So...

y = large

x = small

(It really doesn't matter what variable you use, I just used these)

Now, we know that Estelle fills 3 large boxes and 5 small boxes and had a total of 170 cans in total. We can represent this by writing...

3y + 5x = 170

Then, we know that the boy worker filled 4 large boxes and 4 small boxes and had a total of 184 cans. We can represent this by writing...

4y + 4x = 184

As you can see, this is a system of equations.

3y + 5x = 170

4y + 4x = 184

We want to know how many cans each small box can hold, so we have to find a common number for x since x represents the small box.

To do this, we have to multiply the first equation by 4 and the second by 3. Here's what I mean...

4 (3y + 5x = 170)

3 (4y + 4x = 184)

When we do this you get...

12y + 20x = 680

12y + 12x = 552

Notice how the y's are now both 12. We had to do that in order to get rid of y, they had to equal the same number. Now, subtract...

8x = 128

Divide by 8...

x = 16

That means that...

YOUR ANSWER: Each small box can hold up to 16 cans.

I hope this helps :)

The inverse Laplace transform of 1 / (3s² + 5s + 1) is f(t) = 1/2 × [tex]e^{(-t)[/tex]- 1/2 × [tex]e^{(-t/3)[/tex].

To find the inverse Laplace transform of the function F(s) = 1 / (3s² + 5s + 1), we can use partial fraction decomposition and reference tables for Laplace transforms.

Step 1: Factorize the denominator

Factorize the denominator of the function 3s² + 5s + 1 to find its roots:

3s² + 5s + 1 = (s + 1)(3s + 1)

Step 2: Write the partial fraction decomposition

Write the function F(s) as a sum of partial fractions:

F(s) = A / (s + 1) + B / (3s + 1)

Step 3: Determine the values of A and B

To find the values of A and B, we can multiply both sides of the equation by the common denominator and equate the numerators:

1 = A(3s + 1) + B(s + 1)

Expand the right side and collect like terms:

1 = (3A + B)s + (A + B)

By equating the coefficients of s and the constant terms on both sides, we get a system of equations:

3A + B = 0

A + B = 1

Solving this system of equations, we find A = 1/2 and B = -1/2.

Step 4: Write the inverse Laplace transform

Using the partial fraction decomposition, we can now write the inverse Laplace transform:

F(s) = 1/2 × (1 / (s + 1)) - 1/2 × (1 / (3s + 1))

Referring to Laplace transform tables, we find that the inverse Laplace transform of 1 / (s + a) is [tex]e^{(at)[/tex], and the inverse Laplace transform of 1 / (s - a) is [tex]e^{(at)[/tex]. Therefore, applying these results, we have:

f(t) = 1/2 × [tex]e^{(-t)[/tex] - 1/2 × [tex]e^{(-t/3)[/tex]

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Answer:

6 ft

Step-by-step explanation:

The triangle CDE is half of the size of triangle ABC, so if you match up the corresponding sides, DE is half of AC. AC is 12 so DE is 6.

hope is helped!

The exponential function passing through the points (-3, 1/2) and (3, 32) can be represented by the equation [tex]f(x) = (2^{(5/6)}) * (2^{(x/6)})[/tex].

To find the exponential function passing through the given points, we can start by assuming the general form of an exponential function, [tex]f(x) = a * (b^x)[/tex], where a and b are constants to be determined. Plugging in the coordinates (-3, 1/2) into this equation gives us [tex]1/2 = a * (b^{(-3)})[/tex], and plugging in (3, 32) gives us [tex]32 = a * (b^3)[/tex].

Now, we can solve this system of equations to find the values of a and b. Taking the ratio of the two equations, we get [tex](1/2) / 32 = (a * (b^{(-3)})) / (a * (b^3))[/tex], which simplifies to [tex]1/64 = 1/b^6[/tex]. Solving for b, we find [tex]b = 2^{(1/6)[/tex].

Substituting this value back into either of the original equations, we can solve for a. Using the equation [tex]1/2 = a * (2^{(-3/6)})[/tex], we find [tex]a = 2^{(5/6)[/tex].

Therefore, the exponential function passing through the given points is [tex]f(x) = (2^{(5/6)}) * (2^{(x/6)})[/tex].

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Answer:

The amount invested in ADB is ¢1363.[tex]\overline 3[/tex] =

The amount invested in Barclays is ¢3,427.[tex]\overline 3[/tex]

The amount invested in GCB is ¢1,713.[tex]\overline 3[/tex]

Step-by-step explanation:

The parameters of the investment Stonewall made are;

The amount in interest he receives from ADB, Barclays and GCB = ¢250

The amount of interest ADP pays = 2% per annum

The amount of interest Barclays pays = 4% per annum

The amount of interest GCB pays = 5% per annum

The amount invested in Barclays = The amount invested in ADB and GCB + ¢350

The amount invested in Barclays = 2 × The amount invested in GCB

a) Let 'x', represent the amount invested in ADB, 'y' represent the amount invested in Barclays, and 'z', represent the amount invested in GCB

We have;

y = x + z + 350

y = 2·z

0.02·x + 0.04·y + 0.05·z = 250

Therefore, we get the three linear equations as follows;

-x + y - z = 350...(1)

y - 2·z = 0...(2)

0.02·x + 0.04·y + 0.05·z = 250...(3)

Using Matrix inversion, we have;

[tex]\left[\begin{array}{ccc}-1&1&-1\\0&1&-2\\0.02&0.04&0.05\end{array}\right] \times \left[\begin{array}{c}x&y&z\end{array}\right] = \left[\begin{array}{c}350&0&250\end{array}\right][/tex]

The transpose of the 3 by  3 matrix [tex]M^T[/tex] is given as follows;

[tex]M^T = \left[\begin{array}{ccc}-1&0&0.02\\1&1&0.04\\-1&-2&0.05\end{array}\right][/tex]

The Adjugate Matrix is given as follows;

[tex]Adj = \left[\begin{array}{ccc}0.13&-0.09&-1\\-0.04&-0.03&-2\\-0.02&0.06&-1\end{array}\right][/tex]

The inverse of the matrix = Adj/Det where, Det = -0.15, is therefore;

[tex]M^{-1} = \left[\begin{array}{ccc}\dfrac{-13}{15} &\dfrac{3}{5} &\dfrac{20}{3} \\\\\dfrac{4}{15} &\dfrac{1}{5} &\dfrac{40}{3} \\\\\dfrac{2}{15} &-\dfrac{2}{5} &\dfrac{20}{3} \end{array}\right][/tex]

We therefore, get the solution as follows;

[tex]\left[\begin{array}{ccc}\dfrac{-13}{15} &\dfrac{3}{5} &\dfrac{20}{3} \\\\\dfrac{4}{15} &\dfrac{1}{5} &\dfrac{40}{3} \\\\\dfrac{2}{15} &-\dfrac{2}{5} &\dfrac{20}{3} \end{array}\right]\times \left[\begin{array}{c}350&0&250\end{array}\right] = \left[\begin{array}{c}\dfrac{4,090}{3} \\&\dfrac{10,280}{3} \\ & \dfrac{5,140}{3} \end{array}\right][/tex]

[tex]\begin{array}{c}x = \dfrac{4,090}{3} \\&y = \dfrac{10,280}{3} \\ & z = \dfrac{5,140}{3} \end{array}[/tex]

The amount invested in ADB, x = ¢4,090/3 = ¢1363.[tex]\overline 3[/tex]

The amount invested in Barclays, y = ¢10,282/3 = ¢3,427.[tex]\overline 3[/tex]

The amount invested in GCB, z = ¢5,140/3 = ¢1,713.[tex]\overline 3[/tex]