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LUCKY_DIMON [66]

3 years ago

12

The quadratic function that approximates the height of a javelin throw is h(t) = -0.08t2 +4.48, where t is the time in seconds a

fter it is thrown and h is the javelin's height in feet. How long will it take for the javelin to hit the ground?

1 answer:

Advocard [28]3 years ago

3 0

I believe the answer is .21345

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4x^2 = 92 + 7x solve the equation using the guadratic formula, show all work

katovenus [111]

$4x^2=92+7x\ \ \ \ |-4x^2\\\\-4x^2+7x+92=0\\\\a=-4;\ b=7;\ c=92\\\\b^2-4ac=7^2-4\cdot(-4)\cdot92=49+1472=1521\\\\\sqrt{b^2-4ac}=\sqrt{1521}=39\\\\x_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a};\ x_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\\\\x_1=\dfrac{-7-39}{2\cdot(-4)}=\dfrac{-46}{-8}=\dfrac{23}{4}\\\\x_2=\dfrac{-7+39}{2\cdot(-4)}=\dfrac{32}{-8}=-4$

$\text{Answer:}\ A.\ \left\{\dfrac{23}{4};\ -4\right\}$

4 0

4 years ago

NEED HELP. WILL MARK BRAINLIST AND DON'T SPAM

Verdich [7]

They are both dependent for “win” because “win” is less than the total number of losses.. and the for question B it is independent because it is more points than the number of wins..

7 0

3 years ago

What is the equation of the line that is perpendicular to and has the same y-intercept as the given line? y = One-fifthx + 1 y =

frutty [35]

Answer: tHE ANSWER IS C

Step-by-step explanation: I just took the quiz.

7 0

4 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%20" id="TexFormula1" title=" " alt=" " align="absmiddle" class="latex-formula"><br><img src="

seropon [69]

$\text{The range of}\ y=-3^x\ \text{is}\ (-\infty;\ 0).\\\\\text{The range of}\ f(x)=-3^x+7\ \text{is}\ (-\infty;\ 7)$

6 0

3 years ago

1. Find the vertices and locate the foci for the hyperbola whose equation is given.

Irina18 [472]

$\bf \cfrac{(x-{{ h}})^2}{{{ a}}^2}-\cfrac{(y-{{ k}})^2}{{{ b}}^2}=1
\qquad center\ ({{ h}},{{ k}})\qquad
 vertices\ ({{ h}}\pm a, {{ k}})\\\\
-----------------------------\\\\
\textit{now let's take a look at yours}
\\\\\\
49x2 - 16y2 = 784\implies \cfrac{49x^2}{784}-\cfrac{16y^2}{784}=1
\\\\\\
\cfrac{x^2}{16}-\cfrac{y^2}{49}=1\implies \cfrac{(x-0)^2}{4^2}-\cfrac{(y-0)^2}{7^2}=1
\\\\\\
recall\implies center\ ({{ h}},{{ k}})\qquad vertices\ ({{ h}}\pm a, {{ k}})
\\\\\\
$

$\bf \textit{now, for the foci, the foci are "c" distance from the center point}\\\\\
whereas\qquad c=\sqrt{a^2+b^2}\qquad \textit{ that is }\qquad h\pm \sqrt{a^2+b^2}$

notice your "a" and "b" components, to get the distance "c" from the center to either foci and the vertices, of course, are h + a, k and h - a, k

5 0

3 years ago