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3 years ago

I need help I’m stuck

Mathematics

1 answer:

Svetlanka [38]3 years ago

3 0

The vertex of the angle must stay the same so F must always be in the middle.
Therefore you have EFG
And GFE

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Magan bought an electric bill at 85% of the regular price she paid $32.89 for the drill what was the regular price

Natasha2012 [34]

Answer:

$38.69

Step-by-step explanation:

To find the regular price, create a proportion.

$\frac{32.89}{x}=\frac{85}{100}$

To solve for the regular price x, cross multiply the numerator and denominator of each fraction.

32.89*(100) = x(85)

3289=85x

38.69 = x

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3 years ago

What is 777778978734*79980324768762378945569723

Zielflug [23.3K]

? do you have a photo of your work

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3 years ago

Someone give me songs/artists to listen to

ankoles [38]

Answer:

corpse husband also known as corpse .

Step-by-step explanation:

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2 years ago

Read 2 more answers

6th-grade math, fill out the table! :)

Svetradugi [14.3K]

Answer:

21 wins, table is attached.

Step-by-step explanation:

The ratio of 7 wins to 2 losses is 7:2. A ratio is just saying that every time one value is increased by 7, the other is increased by 2.

So we can fill out the table, in every iteration wins increases by 7 and losses increases by 2.

When we fill this out, we find that when losses is 6, wins is 21, so when you have 2 losses you have 21 wins.

Hope this helped!

4 0

3 years ago

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]

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

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

6 0

3 years ago