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tatuchka [14]
3 years ago
12

How do I do this and solve it

Mathematics
2 answers:
astra-53 [7]3 years ago
7 0

Answer:

7447475047,9659865865,489

Step-by-step explanation:

Nutka1998 [239]3 years ago
4 0
Hey there! :D

Since the sides are equal in length, we know this is an isosceles triangle. It will have two equal angles. 

We already know the vertex is 80 degrees, so the bottom two angles will be equal in measure. All angles in a triangle equal 180 degrees

180-80= 100

100/2= 50 

x= 50

I hope this helps!
~kaikers
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Answer:

\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}=2xy^2\sqrt{x}+2xy\sqrt{y}

Hence, ption B is true.

Step-by-step explanation:

Given the expression

\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}

solving the expression

\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}

as

\sqrt{x^2y^3}=xy\sqrt{y}

2\sqrt{x^3y^4}=2xy^2\sqrt{x}

so the expression becomes

\:\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}=xy\sqrt{y}+2xy^2\sqrt{x}+xy\sqrt{y}

Group like terms

                                        =2xy^2\sqrt{x}+xy\sqrt{y}+xy\sqrt{y}

Add similar elements

                                        =2xy^2\sqrt{x}+2xy\sqrt{y}

Therefore, we conclude that:

\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}=2xy^2\sqrt{x}+2xy\sqrt{y}

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