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2 years ago

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Both Rachel and Dominique throw tennis balls into the air. At any time, t, the height, h, of Rachel’s ball is modeled by the equ

ation h = –16t2 30t 5. Dominique throws his tennis ball with the same acceleration, a, from the same initial height, mc028-1. Jpg, but with an initial velocity, v, double that of Rachel’s. Which equation best models the height of Dominique’s tennis ball? h(t) = at2 vt h0 h = –16t2 30t 10 h = –32t2 60t 10 h = –32t2 30t 5 h = –16t2 60t 5.

1 answer:

WITCHER [35]2 years ago

5 0

67t I know it’s right because I got it right in my test

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Which inequality is shown in this graph? (0, 2) (-1,-2)

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Answer:

D) y = 4x + 2

Step-by-step explanation:

<u><em>Explanation</em></u>:-

Given ( 0, 2) ( -1 , -2 )

Given in equalities y = 4x + 2

put point (0,2) in equation y = 4x +2

2 = 4(0) +2

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Therefore satisfies the point (0,2) equation y = 4x +2

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Therefore satisfies the point (-1,-2) equation y = 4x +2

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3 years ago

Simplify this expression.<br><br> 7+12÷(−3)<br><br> 3<br><br> 7<br><br> 11<br><br> 12

amm1812

Hi there!

»»————-　★　————-««

I believe your answer is:

3

»»————-　★　————-««

Here’s why:

⸻⸻⸻⸻

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3 years ago

What is 8,624 divided by 98

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3 years ago

The diagonal of a screen is 152 cm the length measures 132 cm what is the height

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4 years ago

SOLVE. integration of (1+v^2) /(1-v^3)

s2008m [1.1K]

$\displaystyle\int\frac{1+v^2}{1-v^3}\,\mathrm dv$

$1-v^3=(1-v)(1+v+v^2)$
$\implies\dfrac{1+v^2}{1-v^3}=\dfrac a{1-v}+\dfrac{b_0+b_1v}{1+v+v^2}$
$\implies\dfrac{1+v^2}{1-v^3}=\dfrac{a(1+v+v^2)+(b_0+b_1v)(1-v)}{1-v^3}$
$\implies 1+v^2=(a+b_0)+(a-b_0+b_1)v+(a-b_1)v^2$
$\implies\begin{cases}a+b_0=1\\a-b_0+b_1=0\\a-b_1=1\end{cases}\implies a=\dfrac23,b_0=\dfrac13,b_1=-\dfrac13$

So,

$\displaystyle\int\frac{1+v^2}{1-v^3}\,\mathrm dv=\dfrac23\int\frac{\mathrm dv}{1-v}+\dfrac13\int\frac{1-v}{1+v+v^2}\,\mathrm dv$

The first integral is easy. For the second, since the derivative of the denominator is $(1+v+v^2)=1+2v$ , we can add and subtract $3v$ to get

$\dfrac{1-v}{1+v+v^2}=\dfrac{1+2v-3v}{1+v+v^2}=\dfrac{1+2v}{1+v+v^2}-\dfrac{3v}{1+v+v^2}$

and for the first term employ a substitution. For the remaining term, we can complete the square in the denominator, then use a trigonometric substitution:

$\displaystyle\int\frac{1+2v}{1+v+v^2}\,\mathrm dv=\int\frac{\mathrm dt}t$

where $t=1+v+v^2\implies\mathrm dt=(1+2v)\,\mathrm dv$ , and

$\displaystyle\int\frac{3v}{1+v+v^2}\,\mathrm dv=3\int\frac v{\left(v+\frac12\right)^2+\frac34}\,\mathrm dv$

Then taking $v+\dfrac12=\dfrac{\sqrt3}2\tan s\implies \mathrm dv=\dfrac{\sqrt3}2\sec^2s\,\mathrm ds$ gives

$\displaystyle3\int\frac v{\left(v+\frac12\right)^2+\frac34}\,\mathrm dv=3\int\frac{\frac{\sqrt3}2\tan s-\frac12}{\left(\frac{\sqrt3}2\tan s\right)^2+\frac34}\left(\frac{\sqrt3}2\sec^2s\right)\,\mathrm ds$
$=\displaystyle\sqrt3\int(\sqrt3\tan s-1)\,\mathrm ds$
$=\displaystyle3\int\tan s\,\mathrm ds-\sqrt3\int\mathrm ds$

Now we're ready to wrap up.

$\displaystyle\int\frac{1+v^2}{1-v^3}\,\mathrm dv=\dfrac23\int\frac{\mathrm dv}{1-v}+\dfrac13\int\frac{1-v}{1+v+v^2}\,\mathrm dv$
$=\displaystyle-\frac23\ln|1-v|+\frac13\left(\int\frac{1+2v}{1+v+v^2}\,\mathrm dv-\int\frac{3v}{1+v+v^2}\,\mathrm dv\right)$
$=\displaystyle-\frac23\ln|1-v|+\frac13\int\frac{\mathrm dt}t-\frac13\left(3\int\tan s\,\mathrm ds-\sqrt3\int\mathrm ds\right)$
$=\displaystyle-\frac23\ln|1-v|+\frac13\ln|t|-\int\tan s\,\mathrm ds+\frac1{\sqrt3}\int\mathrm ds$
$=\displaystyle-\frac23\ln|1-v|+\frac13\ln|1+v+v^2|+\ln|\cos s|+\frac s{\sqrt3}+C$
$=\displaystyle-\frac23\ln|1-v|+\frac13\ln|1+v+v^2|+\ln\left|\frac{\sqrt3}{2\sqrt{1+v+v^2}}\right|+\frac1{\sqrt3}\tan^{-1}\frac{2v+1}{\sqrt3}+C$

This can be simplified a bit using some properties of logarithms to obtain

$=\displaystyle-\frac23\ln|1-v|+\frac13\ln(1+v+v^2)+\left(\ln\frac{\sqrt3}2-\frac12\ln(1+v+v^2)\right)+\frac1{\sqrt3}\tan^{-1}\frac{2v+1}{\sqrt3}+C$
$=\displaystyle-\frac23\ln|1-v|-\frac16\ln(1+v+v^2)+\frac1{\sqrt3}\tan^{-1}\frac{2v+1}{\sqrt3}+C$

6 0

3 years ago