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torisob [31]
2 years ago
8

Simplify the following algebraic expressions.

Mathematics
2 answers:
klio [65]2 years ago
7 0

Answer:

A) 8x - 4   ||   B) 6x + 29

Step-by-step explanation:

A) -2x + 5 + 10x - 9

→ -2x+10x+5-9

→ -2x+10x-4

→ 8x - 4

B) 3(x + 7) + 2(-x + 4) + 5x

→ 3x + 21 - 2x + 8 + 5x

→ 6x + 29

labwork [276]2 years ago
4 0

Answer:

A) 8x - 4

B) 6x + 29

Step-by-step explanation:

A) -2x + 5 + 10x - 9 : given

= (10x - 2x) + (5 - 9) : put like terms together

= 8x - 4 : group

B) 3(x + 7) + 2(-x + 4) + 5x : given

= 3x + 21 - 2x + 8 + 5x : expand

= (3x - 2x + 5x) + (21 + 8) : put like terms together

= 6x + 29 : group

Thanks! Have a great day studying!

Answered by : ms115

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The perimeter it a square field is 224. How long is
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8 0
3 years ago
I just need no. a please help me to prove this. ​
OleMash [197]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = π    and         cos A = cos B · cos C

scratchwork:

  A + B + C = π

               A = π - (B + C)

         cos A = cos [π - (B + C)]                              Apply cos

                    = - cos (B + C)                                    Simplify

                    = -(cos B · cos C - sin B · sin C)          Sum Identity

                    = sin B · sin C - cos B · cos C               Simplify

cos B · cos C = sin B · sin C - cos B · cos C               Substitution

2cos B · cos C = sin B · sin C                                        Addition

                     2=\dfrac{\sin B\cdot \sin C}{\cos B \cdot \cos C}                                     Division

                     2 = tan B · tan C

\text{Use the Sum Identity:}\quad \tan(B+C)=\dfrac{\tan B+\tan C}{1-\tan B\cdot \tan C}

<u>Proof LHS → RHS</u>

Given:                              A + B + C = π

Subtraction:                     A = π - (B + C)

Apply tan:                  tan A = tan(π - (B + C))

Simplify:                               = - tan (B + C)

\text{Sum Identity:}\qquad \qquad \qquad =-\bigg(\dfrac{\tan B+\tan C}{1-\tan B\cdot \tan C}\bigg)

Substitution:                        = -(tan B + tan C)/(1 - 2)

Simplify:                               = -(tan B + tan C)/-1

                                            = tan B + tan C

LHS = RHS:   tan B + tan C = tan B + tan C  \checkmark

5 0
3 years ago
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