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Korvikt [17]
2 years ago
10

Can someone please answer this!!

Mathematics
1 answer:
photoshop1234 [79]2 years ago
5 0

Answer:

40

To find the area you multiply base times height

Step-by-step explanation:

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Add. write your answer in simplest form.<br> -4 _/135 + 7 _/15
Zarrin [17]

\huge\quad\qquad\underline{{\sf Answer}}

Let's solve ~

\qquad \sf  \dashrightarrow \:  - 4 \sqrt{135} + 7 \sqrt{15}

\qquad \tt \dashrightarrow \: - 4 \sqrt{5 \times3 \times 3 \times 3}  + 7 \sqrt{15}

\qquad \tt \dashrightarrow \:( - 4 \times 3) \sqrt{5 \times 3}  + 7 \sqrt{15}

\qquad \tt \dashrightarrow \: - 12 \sqrt{15 }  + 7 \sqrt{15}

\qquad \tt \dashrightarrow \: - 5 \sqrt{15}

7 0
2 years ago
Choose the equation that represents the line passing through the point (2, -5) with a slope of -3.
chubhunter [2.5K]
Your answer is D.) y = -3x + 1
7 0
3 years ago
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Whats the key way to resolve reading math problems
Lady bird [3.3K]
I would say read it well and then write down an equation or expression to represent the data so its easier for you to understand and/or solve.
6 0
3 years ago
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Find the total surface area of the net.​
Elanso [62]

Answer:

87.8 square meters

Step-by-step explanation:

First of all, find the area of the three recthangles.They are all 9m long and 3m wide. 9*3 is 27, and since there are three of them, 27*3 is 81.

To find the area of the two triangles, you could do 3*2.6. Normally, you would halve the values because they are triangles but since there's two of them it cancels out.

3*2.6 is 7.8, and adding that to the area of the rectangles gets the equation 81+7.8 which comes out to 87.8 square meters.

3 0
3 years ago
At the beginning of year 1, Sam invests $700 at an annual compound interest
Aleksandr-060686 [28]

at the beginning of year 4, only 3 years have elapsed, the 4th year hasn't started yet, since it's at the beginning, so at the beginning of year 4 we can say only 4-1 years have elapsed.

~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$700\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=\textit{elapsed years}\dotfill &3 \end{cases}

A=700\left( 1 + \frac{0.05}{1} \right)^{1\cdot 3}\implies A = 700(1+0.05)^3\implies A(4)=700(1+0.05)^{4-1} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill A(n)=700(1+0.05)^{n-1}~\hfill

6 0
2 years ago
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