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anzhelika [568]
2 years ago
8

Twenty-two more than three times a number is equal to the difference between 6464 and three times the number. Find the number.

Mathematics
1 answer:
alexgriva [62]2 years ago
8 0

Answer:

7

Step-by-step explanation:

a number: x

3 times a number: 3x

twenty-two more than 3 times a number: 3x + 22

difference between 64 and 3 times the number: 64 - 3x

3x + 22 = 64 - 3x

6x = 42

x = 7

Answer: 7

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John has 1 chicken jake took one how much does john have left
Talja [164]

Answer:

0 because 1-1=0 and that is the answer

3 0
3 years ago
Read 2 more answers
Evaluate -5-15+4+15
defon

Answer:

-1

Step-by-step explanation:

-5-15+4+15

=-20+19

=-1

7 0
3 years ago
Read 2 more answers
84.87 is 115% of what
elena-s [515]
Divide 84.87 by 1.15, and you get 73.8

Then you can check your work by multiplying 73.8 by 1.15 and you get 84.87
6 0
3 years ago
A business buys a computer for $3,000. After 4 years the value of the computer is expected to be $250. The value, v, can be rela
yuradex [85]

You are given two points in the linear function. At time 0 years, the value is $3000. At time 4 years, the value is $250. This means you have points (0, 3000) and (4, 250). You need to find the equation of the line that passes through those two points.

y = mx + b

m = (y2 - y1)/(x2 - x1) = (3000 - 250)/(0 - 4) = 2750/(-4) = -687.5

Use point (0, 3000).

3000 = -687.5(0) + b

b = 3000

The equation is

y = -687.5x + 3000

Since we are using points (t, v) instead of (x, y), we have:

v = -687.5t + 3000

Answer: d. v = -687.50 t + 3,000

6 0
2 years ago
Solve the equation : 8+10^x=1008
Marizza181 [45]
Hi there.

I'm just going to put the answer for now. I might come back and edit my answer if I figure out how to solve this the long way.

All I did was figure out what exponent you raise 10 to. 

10³ = 1000

So, we now have 8 + 10³ = 1008

Edit:

I did it on paper - here it is broken down.

8 + 10^{x} = 1008

Subtract 8 from both sides.

10^{x} = 1000

We know that 10³ = 1000

<em>x</em> = 3

~
7 0
3 years ago
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