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2 years ago

PLEASE HELP! (: i'm really dumb in math, so. Just wanna make sure I do it right.

Mathematics

2 answers:

Mekhanik [1.2K]2 years ago

8 0

1) ADC
2)HGF
3)segment AB and segment ED because they are on different planes

777dan777 [17]2 years ago

6 0

1) ADC
2)BGF i think...
3) idk what skew means
Helped as much as I can right now...

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Can someone tell me the answer

astra-53 [7]

Answer:

-3

Step-by-step explanation:

(-6)/(+2)

-6/+2

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2 years ago

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Anita can skate 3 1/3 miles in 1 hour. How far can she skate in 2 1/2 hours?

irga5000 [103]

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3 years ago

Númer 9 and 10 please help

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9. -1,07, 1/2. 3........it is all I could see

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2 years ago

What is 4/7 times 5/p

Phantasy [73]

Answer:

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Step-by-step explanation:

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2 years ago

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.

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

quick info:

in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².

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2 years ago