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jeka94
4 years ago
10

Elevate 2p - 3r +t for p =10,r= -4 and t = 5

Mathematics
2 answers:
vodka [1.7K]4 years ago
7 0
2.10-3.-4+5=20+12+5=27
ollegr [7]4 years ago
6 0

Answer:   13

Step-by-step explanation:

2x10=20 -(-3-4)+5

20+5 -12=13

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Please help me, it’s asking what the constant rate of change, if it’s possible.
k0ka [10]
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Have an amazing day~
6 0
3 years ago
Which expression is equivalent to 5(4x+3) - 2x
Troyanec [42]

Good evening  

Answer:

  1. 18x + 15
  2. 3(6x + 5)

Step-by-step explanation:

Unfortunately you didn’t provide the expressions to chose from ,however we can get other expressions From this one like this:

5(4x+3) - 2x = 20x + 15 - 2x

                   = (20x - 2x) + 15

                   = 18x + 15

we can also write 18x + 15 = 3×(6x) + 3×(5x) = 3(6x + 5)

then

these two expressions :

  • 18x + 15
  • 3(6x + 5)

are equivalent to 5(4x+3) - 2x

__________________________________________

:)

6 0
4 years ago
Find the value of 4^2 +6 ÷ 2. <br>O 11<br>O 19<br>O 7​
rewona [7]
The correct answer is 19
6 0
3 years ago
Read 2 more answers
Simplify.<br> 1/2 (x – 6) + 3/2 (x + 2)<br> 2x + 6<br> 2x<br> 2x-4<br> 2x + 3
pochemuha
Answer: B 2x

1/2x-3+3/2x+3
7 0
3 years ago
Please help me....Use the Pythagorean identity
RideAnS [48]

Using the Pythagorean identity, the value of the cosine ratio is \cos(\theta_1) =  \frac{84}{85}

<h3>How to determine the cosine ratio?</h3>

The given parameter is:

\sin(\theta_1) = -\frac{13}{85}

By the Pythagorean identity, we have:

\sin^2(\theta_1) + \cos^2(\theta_1) = 1

So, we have:

(-\frac{13}{85})^2 + \cos^2(\theta_1) = 1

This gives

\cos^2(\theta_1) = 1 - (-\frac{13}{85})^2

Evaluate

\cos^2(\theta_1) = 1 - \frac{169}{7225}

Take LCM

\cos^2(\theta_1) = \frac{7225 -169}{7225}

This gives

\cos^2(\theta_1) = \frac{7056}{7225}

Take the square root of both sides

\cos(\theta_1) = \pm \frac{84}{85}

Cosine is positive in the fourth quadrant.

So, we have:

\cos(\theta_1) =  \frac{84}{85}

Hence, the cosine value is \cos(\theta_1) =  \frac{84}{85}

Read more about Pythagorean identity at:

brainly.com/question/1969941

#SPJ1

4 0
3 years ago
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