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MrRa [10]
3 years ago
11

Simplify :Stepwise answer! ​

Mathematics
1 answer:
Free_Kalibri [48]3 years ago
3 0

Answer:

\boxed{ \bold{ \huge{ \boxed{ \sf{ \frac{14 {x}^{2}  - 96}{ {x}^{4} - 13 {x}^{2}   + 36}}}}}}

Step-by-step explanation:

\sf{ \frac{2}{x - 2}  +  \frac{1}{x - 3}  -  \frac{2}{ x + 2}  -  \frac{1}{ x + 3}}

⇒\sf{ \frac{2(x - 3)(x + 3)(x + 2) + (x - 2)(x + 2)(x  + 3) - 2(x - 2)(x - 3)(x + 3) - (x - 2)(x  +  2)(x - 3)}{(x - 2)(x - 3)(x + 2)(x + 3)}}

Use the formula : a² - b² = ( a + b ) ( a - b )

⇒\sf{ \frac{2( {x}^{2}  - 9)(x + 2) + ( {x}^{2} - 4)(x + 3) - (2x - 4)( {x}^{2} - 9) - ( {x}^{2}   - 4)(x - 3) }{( {x}^{2}  - 4)( {x}^{2}  - 9)}}

Distribute 2 through the parentheses

⇒\sf{ \frac{(2 {x}^{2} - 18)(x + 2) + ( {x}^{2}  - 4)(x + 3) - (2x - 4)( {x}^{2}  - 9) - ( {x}^{2}   - 4)(x - 3)}{( {x}^{2}  - 4)( {x}^{2} - 9) }  }

Multiply the algebraic expressions

⇒\sf{ \frac{2 {x}^{3} + 4 {x}^{2}  - 18x - 36 +  {x}^{3}   + 3 {x}^{2} - 4x - 12 - (2 {x}^{3}  - 18x - 4 {x}^{2}   + 36) - ( {x}^{3} - 3 {x}^{2}  - 4x + 12) } {( {x}^{2}  - 4)( {x}^{2}  - 9)} }

When there is a ( - ) in front of an expression, change the sign of each term in the expression

⇒\sf{ \frac{2 {x}^{3} + 4 {x}^{2} - 18x - 36  +  {x}^{3} + 3 {x}^{2}    - 4x - 12 - 2 {x}^{3}   + 18x + 4 {x}^{2}  - 36 -  {x}^{3}  + 3 {x}^{2}  + 4x - 12}{( {x}^{2} - 4)( {x}^{2}  - 9) } }

Since two opposites add up to zero, it would be better to remove them from the expression

⇒\sf{ \frac{4 {x}^{2}  - 36 + 3 {x}^{2}  - 12 + 4 {x}^{2}  - 36 + 3 {x}^{2}   - 12}{( {x}^{2}  - 4)( {x}^{2}  - 9)} }

collect like terms and simplify

⇒\sf{ \frac{14 {x}^{2}  - 48 - 36 - 12}{( {x}^{2} - 4)( {x}^{2} - 9)  } }

⇒\sf{ \frac{14 {x}^{2}  - 84 - 12}{( {x}^{2}  - 4)( {x}^{2} - 9) } }

⇒\sf{ \frac{14 {x}^{2}  - 96}{ ({x}^{2} - 4)( {x}^{2}  - 9)}  }

Multiply : ( x² - 4 ) and ( x² - 9 )

⇒\sf{ \frac{14 {x}^{2}  - 96}{ {x}^{4}  - 13 {x}^{2}  + 36}}

Hope I helped!

Best regards! :D

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Anton [14]

let

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(perpendicular) a = 8 in

(base) b =?

By using Pythagoras theorem

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3 0
3 years ago
In the figure below, points M, P, H, J, and K lie in plane Z.
marishachu [46]

Answers:

  • (a) LN
  • (b) PHJ
  • (c) K
  • (d) HK

===================================================

Explanation:

Part (a)

Lines LN and PN have the point N in common. This is the intersection point.

-----------------

Part (b)

To name a plane, pick any three non-collinear points that are inside it. We cannot pick points H, J, K together because infinitely many planes pass through it. Imagine the piece of flat paper able to rotate around this axis (like a propeller). Having the points not all on the same line guarantees we form exactly one unique plane.

I'll pick the non-collinear points P, H and J to get the name Plane PHJ. Other answers are possible.

------------------

Part (c)

Points H, J and K are collinear as they are on the same line. Pick either H or K to fill out the answer box. I'll go with point K

------------------

Part (d)

Point P and line HK are coplanar. They exist in the same flat plane, or on the same sheet of flat paper together.

We can think of that flat plane as the ground level while something like point N is underground somewhere. So point N and anything on that ground plane wouldn't be coplanar.

Note: there are other possible names for line HK such as line JH or line JK. The order doesn't matter when it comes to naming lines.

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konstantin123 [22]
Use substitution. 

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y = 2 - x -> y = 2 - (-1) -> y = 3
y = 4x + 7 -> y = 4*-1 + 7 -> y = -4 + 7 -> y = 3

So C) is the correct answer.
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ozzi
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So the answer is D

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