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3 years ago

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Solve for x. 8x + 13 12x +7>>

2 answers:

Vladimir79 [104]3 years ago

3 0

ANSWER:

As they are alternate exterior angles, so they must be equal.

8x + 13 = 12x + 7

8x - 12x = 7 - 13

- 4x = - 6

x = 3/2.

Masteriza [31]3 years ago

3 0

Answer:

Solution given:

8x+13+12x+7=180°[exterior angle's sum is supplementary]

20x=180°-20

x=160°/20°

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a volley ball player servers the ball. The ball follows a path given by the equation y=-0.01x^2+0.5x+3 where x and y are measure

dolphi86 [110]

Answer:

(a)

Distance from player should be 13.82 feet or 36.2 feet

(b)

The ball will go over the net

Step-by-step explanation:

we are given

The ball follows a path given by the equation

$y=-0.01x^2+0.5x+3$

where

x and y are measured in feet and the origin is on the court directly below where the player hits the ball

(a)

net height is 8 ft

so, we can set y=8

and then we can solve for x

$8=-0.01x^2+0.5x+3$

$8\cdot \:100=-0.01x^2\cdot \:100+0.5x\cdot \:100+3\cdot \:100$

$800=-x^2+50x+300$

$-x^2+50x-500=0$

$x^2-50x+500=0$

we can use quadratic formula

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-50\pm \sqrt{50^2-4\left(-1\right)\left(-500\right)}}{2\left(-1\right)}$

$x=5\left(5-\sqrt{5}\right),\:x=5\left(5+\sqrt{5}\right)$

$x=13.82,x=36.2$

So, distance from player should be 13.82 feet or 36.2 feet

(b)

we can plug x=30 and check whether y=8 ft

$y=-0.01(30)^2+0.5(30)+3$

$y=9ft$

we know that

height of net is 8 ft

so, the ball will go over the net

5 0

3 years ago

HELP WITH THESE QUESTIONS PART 2.

polet [3.4K]

Answer: (a) IV (b) positive (c) $\frac{\pi}{4}$ (d) C (e) $\sqrt{2}$

<u>Step-by-step explanation:</u>

a) $\frac{15\pi}{4}$ - $\frac{8\pi}{4}$ = $\frac{7\pi}{4}$ , which is located in Quadrant IV <em>per the Unit Circle. </em>

b) sec is $\frac{1}{cos}$ . Cos is the x-coordinate. The x-coordinate inQuadrant IV is positive.

c) the reference angle is the angle from $\frac{7\pi}{4}$ to 2π = $\frac{\pi}{4}$

d) since the angle is below the x-axis and the reference angle is $\frac{\pi}{4}$ , then the angle is equal to $-sec(\frac{\pi}{4}$ )

e) sec = $\frac{1}{cos}$ ⇒ sec $\frac{7\pi}{4}$ = $\frac{2}{\sqrt{2} }$ = $\frac{2}{\sqrt{2} }$ $(\frac{\sqrt{2}} {\sqrt{2}})$ = $\frac{2\sqrt{2} }{2}$ = $\sqrt{2}$

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Answer: (a) $\frac{3\pi}{2}$ (b) (0, -1) (c) A (d) -1

<u>Step-by-step explanation:</u>

a) $\frac{11\pi}{2}$ - $\frac{4\pi}{2}$ = $\frac{7\pi}{2}$ - $\frac{11\pi}{2}$ = $\frac{3\pi}{2}$

b) $\frac{3\pi}{2}$ is on the Unit Circle at (0, -1)

c) sin = $\frac{opposite}{hypotenuse}$ which equals $\frac{y}{r}$ on the Unit Circle.

d) sin is the y-coordinate. sin $(\frac{3\pi}{2})$ = -1

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13 mins left oh gees oh gosh

gavmur [86]

Answer:

I think that it is a quadratic

7 0

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Doss [256]

Each of Danny's friends contributes $12. All you have to do is divide 84/7.

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3 years ago

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Step-by-step explanation:

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