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const2013 [10]
3 years ago
5

Factor completely 20x^3y + 30x^2y^2.

Mathematics
1 answer:
weqwewe [10]3 years ago
6 0

Answer:

10x^2 y(2x + 3y)

Step-by-step explanation:

20x^3 y + 30x^2 y^2.

Factor 10x^2y out of 20x^3y.

10x^2 y (2x) + 30x^2 y^2  

Factor 10x^2y out of 30x^2y^2.

10x^2 y (2x) + 10x^2 y (3y)  

Factor 10x^2y out of 10x^2 y (2x) + 10x^2 y (3y).

10x^2 y(2x + 3y)

you factor out 10x^2y from both side which you will then get 10x^2 y (2x) + 10x^2 y (3y)  than you factor out 10x^2y again and get 10x^2 y(2x + 3y)  your third option

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3 years ago
Let us analyze the following setting. You are given a circle of unit circumference. You pickkpoints on the circle independently
garik1379 [7]

Answer:

We have been given a unit circle which is cut at k different points to produce k different arcs. Now we can see firstly that the sum of lengths of all k arks is equal to the circumference:

\small \sum_{i = 1}^{k} L_i= 2\pi

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7 0
3 years ago
Sin167cos107-cos167sin107
insens350 [35]
Sin 167 = 0.22495

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(0.22495)(-0.29237) - ( -0.97437)(0.95630)

= 0.86602

hope this helps :)



6 0
3 years ago
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