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andriy [413]
4 years ago
14

How to solve the question

Mathematics
2 answers:
Pani-rosa [81]4 years ago
4 0
It should be 90cm
So it says 20cm multiply that by2 which is 40
Then add all the 10.
That is it!
Norma-Jean [14]4 years ago
4 0
Find area of quadrant with radius 10 cm and subtract the triangle. Multiply that by 2 for area at right corner. Let’s call this K
Find area of quadrant with radius 20 cm and subtract a semicircle and square with area of 100 cm^2. Add K to get answer.
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What point of concurrency in a right triangle is also the midpoint of the hypotenuse?
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svp [43]

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2 years ago
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k0ka [10]

We can rewrite the expression under the radical as

\dfrac{81}{16}a^8b^{12}c^{16}=\left(\dfrac32a^2b^3c^4\right)^4

then taking the fourth root, we get

\sqrt[4]{\left(\dfrac32a^2b^3c^4\right)^4}=\left|\dfrac32a^2b^3c^4\right|

Why the absolute value? It's for the same reason that

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since both (-x)^2 and x^2 return the same number x^2, and |x| captures both possibilities. From here, we have

\left|\dfrac32a^2b^3c^4\right|=\left|\dfrac32\right|\left|a^2\right|\left|b^3\right|\left|c^4\right|

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3 0
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