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olganol [36]
3 years ago
9

Solve for u 2(4u-5)=70 Simplify as much as possible

Mathematics
1 answer:
MakcuM [25]3 years ago
8 0
2(4u-5)=70
divide  2 from both sides
4u-5=35
add 5 to both sides
4u=40
divide 4 from both sides
u=10
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3 -14x -x + 25 + 15 simplify​
viktelen [127]

Add the numbers:

3-14x-x+25+15= 43-14-x

Combine like terms:

43-15x

Rearrange terms:

43-15x

-15x+43

Answer:

-15x+43

Hope this helps

4 0
3 years ago
Read 2 more answers
Can someone please help me asappp I’ll mark brainlist !!!
DanielleElmas [232]

Answer:

D. y = 76x

Step-by-step explanation:

First you find the gradient, by using the gradient formula

rise over run

380 - 228

---------------- <- division sign

   5 - 3

152/2 = 76

Hence y = 76x + c

Find c by subbing points in. I am going to use this as an example (5, 380)

380 = 76(5) + c

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5 0
3 years ago
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Evgesh-ka [11]

Answer:

\frac{29}{18}  = 1 \frac{11}{18}

Step-by-step explanation:

To solve this question, we definitely need to make w the subject.

w -  \frac{5}{6}  =  \frac{7}{9}  \\ w -  \frac{5}{6}  +  \frac{5}{6}  =  \frac{7}{9}  +  \frac{5}{6}  \\ w =  \frac{29}{18}  \\  = 1 \frac{11}{18}

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2 years ago
300 bears were out of hibernation. after 2 years, the population grew to 381 bears. if the population continues to grow how many
Nat2105 [25]

Answer:

561 because if you multiple 300 /8 times the number of the before populaiton u got it right Step-by-step explanation:


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3 years ago
True or false:
Gala2k [10]

1. False. f'(c)=0 does not necessarily mean that f(c) is a maximum or minimum. It could just as easily be a saddle point. For example, consider the function f(x)=x^3 with f'(x)=3x^2=0 when x=0, yet f(x) for x and f(x)>0 for x>0.

2. True. If f'(x)=0 for all x in (a,b), then f must be constant on that interval.

3. False. In order for the MVT to apply, f must be continuous on the closed interval. But \dfrac1{x^2} does not exist at x=0.

3 0
4 years ago
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