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2 years ago

In the figure below, m || h. Find the values of z and y.

Mathematics

2 answers:

lawyer [7]2 years ago

7 0

1192828 and also the coordinates 1227

Law Incorporation [45]2 years ago

6 0

Answer:

z = 107, y = 25

Step-by-step explanation:

z and 73 are same- side interior angles and are supplementary, sum to 180°

z + 73 = 180 ( subtract 73 from both sides )

z = 107

z and (3y + 32) are vertical angles and are congruent, then

3y + 32 = 107 ( subtract 32 from both sides )

3y = 75 ( divide both sides by 3 )

y = 25

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2 years ago

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Answer:

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Step-by-step explanation:

step 1

At about what time did the ball reach the maximum?

Let

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t ---> the time in seconds

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

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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

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

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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

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

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

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

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

the positive value is

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

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