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3 years ago

Pls pls help I’m really confused could someone explain to me pls

Mathematics

1 answer:

kenny6666 [7]3 years ago

3 0

Answer: 11.6

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Explanation:

See the attached image for a visual reference

The distance from point D to E is 21 units. Point C is the midpoint, so CD is 10.5 units long (21/2 = 10.5)

We have a right triangle ACD. The legs are
AC = 5
CD = 10.5
The hypotenuse is
AD = x
Because AD is another radius of the same circle

Use the pythagorean theorem to find x
a^2 + b^2 = c^2
5^2 + 10.5^2 = x^2
25 + 110.25 = x^2
135.25 = x^2
x^2 = 135.25
x = sqrt(135.25)
x = 11.629703349613
which rounds to 11.6 when rounding to the nearest tenth (one decimal place)

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Answer:

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Step-by-step explanation:

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3 years ago

What is the product of -3 1/4 aND -1 1/2

Soloha48 [4]

Answer:

Step-by-step explanation:

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8 0

3 years ago

Read 2 more answers

The height h (in feet) of an object dropped from a ledge after x seconds can be modeled by h(x)=−16x2+36 . The object is dropped

kakasveta [241]

Check the picture below.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases}$

so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?

$\bf h(x)=-16x^2+36\implies \stackrel{h(x)}{0}=-16x^2+36\implies 16x^2=36 \\\\\\ x^2=\cfrac{36}{16}\implies x^2 = \cfrac{9}{4}\implies x=\sqrt{\cfrac{9}{4}}\implies x=\cfrac{\sqrt{9}}{\sqrt{4}}\implies x = \cfrac{3}{2}~~\textit{seconds}$

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2 = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill$