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NeTakaya
3 years ago
10

Which systems of equations intersect at point A in this graph?

Mathematics
1 answer:
abruzzese [7]3 years ago
3 0

Answer:

The point of intersection of the system of equations is:

(x, y) =  (-2, 1)

The correct system of equations intersect at point A in this graph will be:

\begin{bmatrix}y=4x+9\\ y=-3x-5\end{bmatrix}

Thus, the second option is correct.

Step-by-step explanation:

Given the point

  • A (-2, 1)

Let us check the system of equations to determine whether it intersect at point A in this graph.

Given the system of equations

\begin{bmatrix}y=4x+9\\ y=-3x-5\end{bmatrix}

Arrange equation variable for elimination

\begin{bmatrix}y-4x=9\\ y+3x=-5\end{bmatrix}

so

y+3x=-5

-

\underline{y-4x=9}

7x=-14

so the system of equations becomes

\begin{bmatrix}y-4x=9\\ 7x=-14\end{bmatrix}

Solve 7x = -14 for x

7x=-14

Divide both sides by 7

\frac{7x}{7}=\frac{-14}{7}

Simplify

x = -2

For y - 4x = 9 plug in x = 2

y-4\left(-2\right)=9

y+4\cdot \:2=9

y+8=9

Subtract 8 from both sides

y+8-8=9-8

Simplify

y = 1

Thus, the solution to the system of equations is:

(x, y) = (-2, 1)

From the attached graph, it is also clear that the system of equations intersects at point x = -2, and y = 1.

In other words, the point of intersection of the system of equations is:

(x, y) =  (-2, 1)

Therefore, the correct system of equations intersect at point A in this graph will be:

\begin{bmatrix}y=4x+9\\ y=-3x-5\end{bmatrix}

Thus, the second option is correct.

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Answer:

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Step-by-step explanation:

Given

<u>Pyramid A</u>

s = 4 -- base sides

V = 36 -- Volume

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s = 6 --- base sides

Required

Determine the volume of pyramid B <em>[Missing from the question]</em>

From the question, we understand that both pyramids are equilateral triangular pyramids.

The volume is calculated as:

V = \frac{1}{3} * B * h

Where B represents the area of the base equilateral triangle, and it is calculated as:

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Where s represents the side lengths

First, we calculate the height of pyramid A

For Pyramid A, the base area is:

B = \frac{1}{2} * s^2 * sin(60)

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B = \frac{1}{2} * 16 * \frac{\sqrt 3}{2}

B = 4\sqrt 3

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V = \frac{1}{3} * B * h

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Make h the subject

h = \frac{3 * 36}{4\sqrt 3}

h = \frac{3 * 9}{\sqrt 3}

h = \frac{27}{\sqrt 3}

To calculate the volume of pyramid B, we make use of:

V = \frac{1}{3} * B * h

Since the heights of both pyramids are the same, we can make use of:

h = \frac{27}{\sqrt 3}

The base area B, is then calculated as:

B = \frac{1}{2} * s^2 * sin(60)

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So:

B = \frac{1}{2} * 6^2 * sin(60)

B = \frac{1}{2} * 36 * \frac{\sqrt 3}{2}

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So:

V = \frac{1}{3} * B * h

Where

B = 9\sqrt 3 and h = \frac{27}{\sqrt 3}

V = \frac{1}{3} * 9\sqrt 3 * \frac{27}{\sqrt 3}

V = \frac{1}{3} * 9 * 27

V = 81

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