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konstantin123 [22]
2 years ago
10

Kayson is looking at two buildings, building A and building B, at an angle of elevation of 59°. Building A is 40 feet away, and

building B is 60 feet away. Which building is taller and by approximately how many feet?
Building A; it is around 66.57 feet taller than building B
Building A; it is around 33.29 feet taller than building B
Building B; it is around 66.57 feet taller than building A
Building B; it is around 33.29 feet taller than building A
Mathematics
1 answer:
Snezhnost [94]2 years ago
4 0

Answer:

last one Building B; it is around 33.29 feet taller than building A Step-by-step explanation:

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In the morning Mary walked 2 1/4 miles around the park. In the afternoon she walked another 5 7/8. how many miles in total did s
anygoal [31]

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8 1/8 total miles walked.

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I like to make fractions simple so I would do this,

9/4 plus 47/8

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3 years ago
Find the slope of the line through the points (7,4) and (-10,4).
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0

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7 0
3 years ago
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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.

4 0
3 years ago
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