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djverab [1.8K]
3 years ago
14

Find the area of each. Use

Mathematics
1 answer:
Oksana_A [137]3 years ago
4 0

Answer:

1. A ≈ 452.39

2. A ≈ 615.75

3. i cant see this one

4. A ≈ 380.13

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Assuming we are rounding to the nearest 100, five numbers that could be rounded up to 600 are:
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And five numbers that could be rounded down to 600 are:
601, 610, 625, 630, 649
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Do you believe that the US should convert to the metric system explain why or why notâ
Anna71 [15]

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Step-by-step explanation:

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2 years ago
If f(x) = 6x +5 and g(x) = 2x – 7 find (fºg)(x)
adell [148]

Answer:

f(g(x)) = 12x - 37

Step-by-step explanation:

Step 1: Define

f(x) = 6x + 5

g(x) = 2x - 7

Step 2: Find f(g(x))

f(g(x)) = 6(2x - 7) + 5

f(g(x)) = 12x - 42 + 5

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7 0
3 years ago
Please help solve this system of equations
stepan [7]

Make a substitution:

\begin{cases}u=2x+y\\v=2x-y\end{cases}

Then the system becomes

\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}

Simplifying the equations gives

\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}

which is to say,

\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}

\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81

\implies\dfrac uv=\pm27

\implies u=\pm27v

Substituting this into the new system gives

\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1

\implies u=\pm27

Then

\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13

(meaning two solutions are (7, 13) and (-7, -13))

8 0
2 years ago
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