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stepladder [879]
2 years ago
11

7 divided by 4 x + 3 Y + 4 divided / 4 x minus 3 Y is equal to 5 / 4

Mathematics
1 answer:
DerKrebs [107]2 years ago
7 0

9514 1404 393

Answer:

  (x, y) = (4, 4)

Step-by-step explanation:

If you attempt to solve this by clearing fractions, you end up with an extraneous solution. Here, we'll solve the linear equations ...

  7a +4b = 5/4

  8b -14a = 3/2

where a = 1/(4x+3y) and b = 1/(4x-3y)

Dividing the second equation by 2 and adding the first, we have ...

  (7a +4b) +1/2(8b -14a) = (5/4) +1/2(3/2)

  8b = 8/4

  b = 1/4

Substituting into the first equation gives ...

  7a +4/4 = 5/4

  a = 1/28

__

Now, we can get back to solving for x and y.

  4x +3y = 1/a = 28

  4x -3y = 1/b = 4

  8x = 32 . . . . . . . . . add the two equations

  x = 4

  3y = 28 -4x = 28 -4(4) = 12 . . . . . use x in the equation for 'a'; rearrange

  y = 4 . . . . divide by 3

The solution to the system of equations is (x, y) = (4, 4).

_____

<em>Additional comment</em>

If you graph these equations, you find they describe hyperbolas that intersect at (0, 0) and (4, 4). The "solution" (0, 0) is extraneous, as both equations are undefined there.

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Answer:

part 1) 0.78 seconds

part 2) 1.74 seconds

Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

h ----> the height of a ball in feet

t ---> the time in seconds

we have

h(t)=-16t^{2}+25t+5

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

so

The x-coordinate of the vertex represent the time when the ball reach the maximum

Find the vertex

Convert the equation in vertex form

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h(t)=-16(t^{2}-\frac{25}{16}t)+5

Complete the square

h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+5+\frac{625}{64}

h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}\\h(t)=-16(t^{2}-\frac{25}{16}t+\frac{625}{1,024})+\frac{945}{64}

Rewrite as perfect squares

h(t)=-16(t-\frac{25}{32})^{2}+\frac{945}{64}

The vertex is the point (\frac{25}{32},\frac{945}{64})

therefore

The time when the ball reach the maximum is 25/32 sec or 0.78 sec

step 2

At about what time did the ball reach the minimum?

we know that

The ball reach the minimum when the the ball reach the ground (h=0)

For h=0

0=-16(t-\frac{25}{32})^{2}+\frac{945}{64}

16(t-\frac{25}{32})^{2}=\frac{945}{64}

(t-\frac{25}{32})^{2}=\frac{945}{1,024}

square root both sides

(t-\frac{25}{32})=\pm\frac{\sqrt{945}}{32}

t=\pm\frac{\sqrt{945}}{32}+\frac{25}{32}

the positive value is

t=\frac{\sqrt{945}}{32}+\frac{25}{32}=1.74\ sec

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