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BlackZzzverrR [31]
3 years ago
8

Simplify (4ab^2)^3 Any help is appreciated

Mathematics
1 answer:
Pie3 years ago
3 0

Answer:

64a^{3} b^{6}

Step-by-step explanation:

1) Distribute the exponent outside of the parentheses to the terms inside the parentheses. In this case, multiply the 3 outside the parentheses with the exponent of each of the terms inside:

(4ab^{2} )^3

4^{3} · a^{3} · b^{6}

2) The variables are fine - you can't simplify them further. All you need to do now is multiply the 4^{3} out. Remember that it's kind of like asking, "what is 4 × 4 ×4?"

4 · 4 · 4 · a^{3} · b^{6}

16 · 4 · a^{3} · b^{6}

64a^{3} b^{6}

Therefore, 64a^{3} b^{6} is the answer.

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Step-by-step explanation:

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Solve the triangle m&lt;N= 118°, m&lt;P= 33° and m=15. Round to the nearest tenth. Please show work.​
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Step-by-step explanation:

M = 180 -33-118 = 29

We can use the rule of sines

sin A          sin B           sin C

------------ = ---------- = ------------

a                  b               c

sin 118          sin 29        

------------ = ----------

n                  15              

Using cross products

15 sin 118 = n sin 29

Divide by sin 29

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sin 33          sin 29        

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p                  15              

Using cross products

15 sin 33 = p sin 29

Divide by sin 29

15 sin 33 / sin 29 = p

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3 0
3 years ago
which of the following gives an equation of a line that passes through the point (6 over 5, -19 over 5) and is parallel to the l
victus00 [196]
One equation for this would be

y = \frac{41}{16} x-\frac{55}{8}

We start by finding the slope between the two points:

m=\frac{y_2-y_1}{x_2-x_1}=\frac{-12-\frac{-19}{5}}{-2-\frac{6}{5}}&#10;\\&#10;\\=(-12+\frac{19}{5}) \div (-2-\frac{6}{5})&#10;\\&#10;\\=(\frac{-60}{5}+\frac{19}{5}) \div (\frac{-10}{5}-\frac{6}{5})&#10;\\&#10;\\=\frac{-41}{5} \div \frac{-16}{5}=\frac{-41}{5} \times \frac{-5}{16}=\frac{41}{16}

A line parallel to this one will have the same slope.  We will use point-slope form to write our equation:

y-y_1=m(x-x_1)&#10;\\&#10;\\y-\frac{-19}{5}=\frac{41}{16}(x-\frac{6}{5})&#10;\\&#10;\\y+\frac{19}{5}=\frac{41}{16}x- \frac{41}{16} \times \frac{6}{5}&#10;\\&#10;\\y+\frac{19}{5}=\frac{41}{16}x-\frac{246}{80}&#10;\\&#10;\\y+\frac{304}{80}=\frac{41}{16}x-\frac{246}{80}&#10;\\&#10;\\y=\frac{41}{16}x-\frac{246}{80}-\frac{304}{80}&#10;\\&#10;\\y=\frac{41}{16}x-\frac{550}{80}&#10;\\&#10;\\y=\frac{41}{16}x-\frac{55}{8}
6 0
3 years ago
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