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3 years ago

7x + 9 = 10 - x please help meeeeeeee

Mathematics

2 answers:

nikklg [1K]3 years ago

8 0

Answer:

$7x + 9 = 10 - x \\ (7 + 1)x = 10 - 9 \\ 8x = 1 \\ x = \frac{1}{8}$

Evgen [1.6K]3 years ago

6 0

Answer - x = 1/8

A brainliest for the answer will be apprecited

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Which expression is equivalent to 4 x 4 x 4 x 5 x 5 ?

ivann1987 [24]

Answer:

4³×5² or 4³5²

Step-by-step explanation:

This is because there are three consecutive 4s, so you have to put the 4 into exponential form 3 times. So the exponential form for the 4s will be 4³. There are also 2 consecutive 5s so you have to include 2 in the exponential form. So, the exponential form of the 5s is 5². Now you have to somehow multiply them together. You could do this by putting a multiplication sign, (4³×5²) or you could bunch them together and it will mean the same thing (4³5²).

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3 years ago

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algol [13]

Answer:

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Step-by-step explanation:

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Evaluate the expression 2(x-5) - 3(x-6)

Nataly_w [17]

Answer:

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Step-by-step explanation:

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3 years ago

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Insert one pair of bracket to make this calculation correct .<br> 7 - 5 - 3 + 4 = 9

trasher [3.6K]

7-(5-3)+4=9 is the answer

3 0

3 years ago

$20000 is invested in an account that earned 6% p.A. Compounding yearly for 3 years. The interest rate then went up to 8% p.A. F

GuDViN [60]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for the first 3 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$20000\\ r=rate\to 6\%\to \frac{6}{100}\dotfill &0.06\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=years\dotfill &3 \end{cases}$

$\bf A=20000\left(1+\frac{0.06}{1}\right)^{1\cdot 3}\implies A=20000(1.06)^3\implies \boxed{A=2382.032} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~ \textit{Compound Interest Earned Amount \underline{for the next 4 years}}$

$\bf A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$2382.032\\ r=rate\to 8\%\to \frac{8}{100}\dotfill &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per annum, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases}$

$\bf A=2382.032\left(1+\frac{0.08}{1}\right)^{1\cdot 4}\implies A=2382.032(1.08)^4\implies \boxed{A\approx 3240.73} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{amount for this period}}{2382.032+3240.73}\implies 5622.762$

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3 years ago