Help fast. do first question only and please show work thx.

Mathematics

1 answer:

amm18122 years ago

7 0

Y = -0.01x^2 + 0.5x + 3

when the height y = 8
-0.01x^2 + 0.5x + 3 = 8
-0.01x^2 + 0.5x - 5 = 0

x = 13.82 and x = 36.18 so the ball is at least 8 ft high from 13.82 and 36.18 feet inclusive.- from the player.

If the player is 30 ft from the net then the ball is likely to go over the net.

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Need help with trigonometry

vovangra [49]

Answer:

$\sf \tan(A) & =\dfrac{9\sqrt{22}}{44}$

Step-by-step explanation:

If angle C is the right angle, then side c is the hypotenuse.

Use Pythagoras' Theorem $a^2+b^2=c^2$ to find the length of side a:

Given:

b = 2√22
c = 13

$\implies a^2+(2 \sqrt{22})^2=13^2$

$\implies a^2+88=169$

$\implies a^2=81$

$\implies a=\sqrt{81}$

$\implies a=9$

<u>Tan Trig Ratio</u>

$\sf \tan(\theta)=\dfrac{O}{A}$

where:

$\theta$ is the angle
O is the side opposite the angle
A is the side adjacent the angle

Given:

$\theta$ = A
O = side opposite angle A = a = 9
A = side adjacent angle A = b = 2√22

$\begin{aligned}\implies \sf \tan(A) & =\dfrac{9}{2\sqrt{22}}\\\\ & =\dfrac{9}{2\sqrt{22}} \times \dfrac{\sqrt{22}}{\sqrt{22}}\\\\ & = \dfrac{9\sqrt{22}}{44} \end{aligned}$