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amm1812
2 years ago
8

CAN SOMEONE PLEASE ANSWER THIS QUICK I HAVE 5 MINS PLS

Mathematics
1 answer:
alex41 [277]2 years ago
6 0

Answer:

$20

Step-by-step explanation:

49 is the outlier

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p_X(x)\geq0

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Write an equation of the line that passes through (4,-6) and is parallel to the line y=-3x-9
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2 years ago
The height of a ball thrown into the air after t seconds have elapsed is h = −16t2 + 40t + 6 feet. What is the first time, t, wh
sp2606 [1]
<h3>The first time when the ball will reach a height of 20 feet is 0.42 seconds</h3>

<em><u>Solution:</u></em>

Given that,

<em><u>The height of a ball thrown into the air after t seconds have elapsed is:</u></em>

h = -16t^2 + 40t + 6

<em><u>What is the first time, t, when the ball will reach a height of 20 feet?</u></em>

Substitute h = 20

20 = -16t^2 + 40t + 6\\\\-16t^2 + 40t + 6 -20 = 0\\\\-16t^2 + 40t -14 = 0\\\\16t^2 -40t + 14 = 0\\\\8t^2 -20t + 7=0

<em><u>Solve by quadractic formula</u></em>

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}

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

\mathrm{For\:}\quad a=8,\:b=-20,\:c=7

t = \frac{-\left(-20\right)\pm \sqrt{\left(-20\right)^2-4\cdot \:8\cdot \:7}}{2\cdot \:8}\\\\t = \frac{20 \pm \sqrt{176}}{16}\\\\t = \frac{20 \pm 4\sqrt{11}}{16}\\\\t = \frac{ 5 \pm \sqrt{11}}{4}\\\\We\ have\ two\ solutions\\\\ t=2.07915, \:t=0.42084

Rounding off we get,

t = 2.08 , t = 0.42

Thus the first time when the ball will reach a height of 20 feet is 0.42 seconds

4 0
3 years ago
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