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Sergeeva-Olga [200]
3 years ago
7

Last Week Kellie ran 28 miles. This week she ran 32 miles. What was the percent change in the number of miles she ran? Please ex

plain.
Mathematics
2 answers:
Archy [21]3 years ago
6 0
4/28= 1/7 = 14.3%...............
Ivenika [448]3 years ago
3 0
Lets say 28 miles is a hurend percent well now she ran 4 more miles so then you dived  4 by 28 and add it to a hurend
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1/2(n-5)=1/2(23-5) I cannot find what’s half of 23-5
liraira [26]

Answer:

n = 23

Step-by-step explanation:

if 1/2 (n-5) = 1/2 (23-5)

that must mean that n=23.

and if you are just looking for 1/2 of 23-5 that is just 9. (23-5) = 18 and half of 18 is 9

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3 years ago
Math work pls help :)​
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8 0
2 years ago
What does 144 =to for a squared
Dafna11 [192]
12 *12 = 144

Therefore 12^2 = 144
5 0
3 years ago
Read 2 more answers
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
2. Factorise: a - b - a + b2.square ​
marshall27 [118]
A - a removes both a’s from equation, then you’re left with b + bsquared, since they are unknown values you cannot combine them, even if it is the same letter, you can only add and subtract unknown values with their same letters together.
7 0
2 years ago
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