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leonid [27]
3 years ago
12

Solve for x pls and thank you

Mathematics
1 answer:
tino4ka555 [31]3 years ago
4 0

Answer:

x= 12

Step-by-step explanation:

∠QPR +∠PQR= ∠HRQ (ext. ∠ of △PQR)

30° +(4x +2)°= (8 +6x)°

32 +4x= 8 +6x <em>(simplify)</em>

<em>Bring x terms on 1 side, constants to the other:</em>

6x -4x= 32 -8

2x= 24 <em>(</em><em>simplify</em><em>)</em>

x= 24 ÷2 <em> (÷2 on both sides)</em>

x= 12

<u>Method</u><u> </u><u>2</u><u>:</u>

∠PRQ= 180° -30° -(4x +2)° (∠ sum of △PQR)

∠PRQ= (148 -4x)°

∠PRQ +∠HRQ= 180° (adj. ∠s on a str. line)

(148 -4x)° +(8 +6x)°= 180°

148 -4x + 8 +6x= 180

156 +2x= 180

2x= 180 -156

2x= 24

x= 24 ÷2

x= 12

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Y = -8/3x + 2...slope here is -8/3. A perpendicular line will have a negative reciprocal slope. To find the negative reciprocal of the slope -8/3, u just flip the slope and change the sign. So our perpendicular line will have a slope of 3/8.

y = mx + b
slope(m) = 3/8
(32,6)...x = 32 and y = 6
now we sub and find b, the y int
6 = 3/8(32) + b
6 = 12 + b
6 - 12 = b
-6 = b

so ur perpendicular equation is : y = 3/8x - 6
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24<br> ,<br> Factor:<br> x²-9<br> A<br> (x + 3XX-3)<br> B<br> (x+3)2<br> с<br> (x-3)<br> (x+3)(x+3)
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Answer:

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Step-by-step explanation:

We are given the expression \displaystyle \large{x^2-9}:—

To factor this expression, we have a formula for it which is <u>difference of two squares</u>:—

\displaystyle \large{a^2-b^2=(a+b)(a-b)}

You can also swap from \displaystyle \large{(a+b)(a-b)} to \displaystyle \large{(a-b)(a+b)} via multiplication property.

From the expression, factor using the formula above:—

\displaystyle \large{x^2-9=(x^2)-(3)^2}\\\displaystyle \large{x^2-9=(x-3)(x+3)}

Therefore, the factored expression is:—

\displaystyle \large{\boxed{(x-3)(x+3)}}

__________________________________________________________

If you have any questions regarding the problem or need clarification of my answer/explanation, do not hesitate to ask in comment!

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