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Dominik [7]
3 years ago
11

Graph the linear equation by using Slope intercepts form

Mathematics
1 answer:
LUCKY_DIMON [66]3 years ago
4 0
Here is for solving for x

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If a transversal crosses three non-parallel lines, how many pairs of vertical angles are formed? Explain your answer. its a writ
Lina20 [59]

Step-by-step explanation:

Each transversal crossing a line generates 2 pairs of vertical angles and for 3 transversal crossing the same line it will be 3 x2 = 6 pairs (or 12 angles)

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Multiplication Property of Equality: They multiplied both sides by 5/3 to isolate x on the left

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8 0
3 years ago
HELP PLEASE!! I DONT UNDERSTAND!!!!!!!!!! THANKS SO MUCH
8090 [49]

Hello, please consider the following.

We will multiply the numerator and denominator by

4+\sqrt{6x}

to get rid of the root in the denominator.

First of all, we cannot divide by 0, right? So, we need to make sure that the denominator is different from 0.

4-\sqrt{6x} =0\sqrt{6x}=4\\\\\text{Take the square}\\\\6x=4^2=16\\\\x=\dfrac{16}{6}=\dfrac{8}{3}

We need to take any x real number different from 8/3 then and simplify the expression.

Let's do it!

\begin{aligned}\dfrac{4}{4-\sqrt{6x}}&=\dfrac{4(4+\sqrt{6x})}{(4+\sqrt{6x})(4-\sqrt{6x})}\\\\&=\dfrac{4(4+\sqrt{6x})}{(4^2-\sqrt{6x}^2)}\\\\&=\dfrac{4(4+\sqrt{6x})}{(16-6x)}\\\\&=\dfrac{2(4+\sqrt{6x})}{(8-3x)}\\\\&\large \boxed{=\dfrac{8+2\sqrt{6x}}{8-3x}}\end{aligned}

Thank you

8 0
3 years ago
Use the identity (x^2+y^2)^2=(x^2−y^2)^2+(2xy)^2 to determine the sum of the squares of two numbers if the difference of the squ
marissa [1.9K]

Answer:

Step-by-step explanation:

(x^2+y^2)^2=(x^2)^2+2x^2y^2+(y^2)^2

Adding and substracting 2x^2y^2

We get

(x^2+y^2)^2=(x^2)^2+2x^2y^2+(y^2)^2 +2x^2y^2-2x^2y^2

And we know a^2-2ab+b^2=(a-b)^2

So we identify (x^2)^2 as a^2 ,(y^2)^2 as b^2 and -2x^2y^2 as - 2ab. So we can rewrite (x^2+y^2)^2=(x^2 - y^2)^2 + 2x^2y^2 + 2x^2y^2= (x^2 - y^2)^2+4x^2y^2= (x^2 - y^2)^2+2^2x^2y^2

Moreever we know (a·b·c)^2=a^2·b^2·c^2 than means 2^2x^2y^2=(2x·y)^2

And (x^2+y^2)^2=(x^2 - y^2)^2 + (2x·y)^2

5 0
3 years ago
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