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Paha777 [63]
3 years ago
9

Scientists are studying the temperature on a distant planet. They find that the surface temperature at one location is 40° Celsi

us. They also find that the temperature decreases by 3° Celsius for each kilometer you go up from the surface.
Let T represent the temperature (in Celsius), and let H be the height above the surface (in kilometers). Write an equation relating T to H,
Mathematics
1 answer:
jok3333 [9.3K]3 years ago
4 0

Answer:

T(h) = 20 - 3h

Step-by-step explanation:

We are told that for every km you go up, the temp goes down 3°.

Let's say you go up 1km, then the temp should be 17°. We can easily represent it by the equation: T(h) = 20 - 3h

If we go up 1km: T(1) = 20 - 3*(1) = 20 - 3 = 17°

If we go up 2km: T(2) = 20 - 3*(2) = 20 - 6 = 14°

If we go up 3km: T(3) = 20 - 3*(3) = 20 - 9 = 11°

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3 years ago
Will give brainliest-H(t)=-16t^2+28t
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3 years ago
Giving brainliest! This is very urgent :))
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Answer:

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5 0
3 years ago
Read 2 more answers
Y is directly proportional to square root of x<br> If y=56 when x=49 find,<br> y when x=81
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\qquad \qquad \textit{direct proportional variation} \\\\ \textit{\underline{y} varies directly with \underline{x}}\qquad \qquad \stackrel{\textit{constant of variation}}{y=\stackrel{\downarrow }{k}x~\hfill } \\\\ \textit{\underline{x} varies directly with }\underline{z^5}\qquad \qquad \stackrel{\textit{constant of variation}}{x=\stackrel{\downarrow }{k}z^5~\hfill } \\\\[-0.35em] \rule{34em}{0.25pt}

\stackrel{\begin{array}{llll} \textit{"y" directly}\\ \textit{proportional to }\sqrt{x} \end{array}}{y = k\sqrt{x}}\qquad \textit{we know that} \begin{cases} y = 56\\ x = 49 \end{cases}\implies 56=k\sqrt{49} \\\\\\ 56=7k\implies \cfrac{56}{7}=k\implies 8=k~\hfill \boxed{y=8\sqrt{x}} \\\\\\ \textit{when x = 81, what is "y"?}\hfill y=8\sqrt{81}\implies y=8(9)\implies y=72

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2 years ago
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Answer:

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8/20 divided by 4 equals 1/5

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