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slamgirl [31]
3 years ago
6

Solve for the variable m. m-n=5

Mathematics
1 answer:
dolphi86 [110]3 years ago
6 0

if you wanna solve for m you have get m by him self.

m-n = 5

m-n+n = 5+n

m = 5+ n

there is your answer. hope it is heplful

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erik [133]
THE ANSWER WILL BE c
7 0
3 years ago
The mean of five numbers is 8. when another number is added the mean is 7. find the number added
sasho [114]

Answer:

\large \boxed{\sf \ \ 2  \ \ }

Step-by-step explanation:

Hello,

<u>The mean of five numbers is 8</u> so we can write

\dfrac{x_1+x_2+x_3+x_4+x_5}{5}=8

<u>When another number is added the mean is 7</u>, let s note x the another number we can write

\dfrac{x_1+x_2+x_3+x_4+x_5+x}{6}=7

From the first equation we can say

x_1+x_2+x_3+x_4+x_5=8*5=40

So the second equation becomes

\dfrac{x_1+x_2+x_3+x_4+x_5+x}{6}=7\\\\ \dfrac{40+x}{6}=7\\\\40 + x = 6*7=42\\\\ x = 42-40 = 2\\

The solution is then 2

Hope this helps

4 0
3 years ago
What is m&lt;1 (please help its due in an hour)
matrenka [14]

Answer:

47...........

hahaha hope it it's right

6 0
3 years ago
liam deposits $2400 into an account that earns 8.3% interest compounded quarterly. how much money will liam have in 5 years.
Harlamova29_29 [7]

Answer:

Total = 2,400 * (1 + (.083/4))^4*5

Total = 2,400 * (1.02075)^20

Total = 2,400 * 1.5079528829

Total = 3,619.09

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Find an equation perpendicular to x=1 and passing through (8,-9)
valina [46]

Answer:

y=-9

Step-by-step explanation:

We want to find an equation of a line that's perpendicular to x=1 that also passes through the point (8,-9).

Note that x=1 is a <em>vertical line </em>since x is 1 no matter what y is.

This means that if our new line is perpendicular to the old, then it must be a <em>horizontal line</em>.

So, since we have a horizontal line, then our equation must be our y-value of our point.

Our y-coordinate of our point (8,-9) is -9.

Therefore, our equation is:

y=-9

And this is in standard form.

And we're done!

3 0
3 years ago
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