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Art [367]
4 years ago
5

Three consecutive odf integers have a sum of 39

Mathematics
2 answers:
Anestetic [448]4 years ago
7 0
11, 13, 15
            ...............
professor190 [17]4 years ago
4 0
The integers: x, x+2, x+4
x+x+2+x+4=39
3x+6=39
3x=33
x=11

the integers are 11, 13, 15
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Write an equation that you can use to solve for x.<br><br> Enter your answer in the box.
Talja [164]

Answer:

x+63 = 90

x = 27

Step-by-step explanation:

The two angles are complementary, so they add to 90 degrees

x+63 = 90

Subtract 63 from each side

x+63-63 = 90-63

x = 27

7 0
3 years ago
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Lemur [1.5K]
Because what please elaborate
4 0
3 years ago
Here is a picture of a math problem.
MAXImum [283]
The coordinates are (6,-9) (-12,-3) (0,-7)

you put 6 in the place of x to get 6+3y=-21 then you must take 6 away from each side to get 3y= -27 then divide each side by 3 so -27 divided by 3 is -9. do this for all of them! hope it helped!
3 0
3 years ago
22.
Vlada [557]

Step-by-step explanation:

\sqrt{p}  =  \sqrt[r]{w -  {as}^{2} }

Find raise each side of the expression to the power of r

That's

<h2>( \sqrt{p} )^{r}  =  (\sqrt[r]{w -  {as}^{2} } ) ^{r}</h2>

we have

<h2>( \sqrt{p} )^{r}  =  w -  {as}^{2}</h2>

Send w to the left of the equation

<h2>( \sqrt{p} )^{r}  - w =  -{as}^{2}</h2>

Divide both sides by - a

We have

<h2>{s}^{2}  =  -\frac{( \sqrt{p} )^{r}  - w}{a}</h2>

Find the square root of both sides

We have the final answer as

<h2>s =  \sqrt{ -\frac{( \sqrt{p} )^{r} - w }{a} } </h2>

Hope this helps you

3 0
3 years ago
8b-ab=7a, .subtracted from 3a-9ab+b<br>​
Tresset [83]

Answer: You can't. Read explanation.

Step-by-step explanation:

You can't subtract an expression from an equation. If you said something like subtract 8b-ab+7a from 3a-9ab+b that would work, but not here.

Let's just assume You mean it as 8b-ab-7a, then (3a-9ab+b)-(8b-ab-7a) = -8ab + 10a - 7b.

Hope that helped,

-sirswagger21

3 0
3 years ago
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