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slavikrds [6]
4 years ago
8

For one month Siera calculated her home town's average high temperature in degrees Fahrenheit. She wants to convert that tempera

ture from degrees Fahrenheit to degrees Celsius using the function C(f)=5/9(f-32) . What does C(F) represent?
Mathematics
1 answer:
Strike441 [17]4 years ago
5 0

Answer:

5

Step-by-step explanation:

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3 years ago
UHHHHH IM GONNA FAIL MATH YALL<br><br> Just answer it ig-?<br> I WILL MARK AS BRAINLIEST
tankabanditka [31]

Answer:

30.96 m^2

Step-by-step explanation:

Side length of the square is also the diameter of the semicircle so:

r = 0.5 * 12 m = 6 m

Since you want the area of the shaded area, you have to subtract the area of both semi circles from the area of the square:

Square Area - Semi Circle Area - Semi Circle Area = Shaded Region Area

(12 m * 12 m) - (0.5 * 3.14 * 6^2 m) - (0.5 * 3.14 * 6^2 m)

= 144 - 56.52 - 56.52

= 30.96 m^2

7 0
3 years ago
heavy rain in oxford caused the creek to rise. the creek rose 3 inches the first day, and each day twice as much as the previous
Fittoniya [83]

48 inches???

Step-by-step explanation:

4 0
3 years ago
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Can someone plz help me with this one problem plzzzzz!!!<br><br> (I’M MARKING BRAINLIEST)!
liraira [26]

Answer:

1:5

5:9

7:11

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I think this is right in not sure though

8 0
3 years ago
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You decide to put $5000 in a savings account to save $6000 down payment on a new car. If the account has an interest rate of 7%
bezimeni [28]

\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill&\$6000\\ P=\textit{original amount deposited}\dotfill &\$5000\\ r=rate\to 7\%\to \frac{7}{100}\dotfill &0.07\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years \end{cases}

\bf 6000=5000\left(1+\frac{0.07}{12}\right)^{12\cdot t}\implies \cfrac{6000}{5000}\approx (1.0058)^{12t}\implies \cfrac{6}{5}\approx(1.0058)^{12t} \\\\\\ \log\left( \cfrac{6}{5} \right)\approx \log[(1.0058)^{12t}]\implies \log\left( \cfrac{6}{5} \right)\approx 12t\log(1.0058) \\\\\\ \cfrac{\log\left( \frac{6}{5} \right)}{12\log(1.0058)}\approx t\implies 2.63\approx t\impliedby \textit{about 2 years, 7 months and 16 days}

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4 years ago
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