Ask question

Ask question

All categories

3 years ago

10

Find three consecutive even integers such that six times the sum of the first and the third is twenty - four greater than eleven

times the second .

1 answer:

dedylja [7]3 years ago

8 0

Answer:

1st = 22

2nd = 24

3rd = 26

Step-by-step explanation:

1st = x

2nd = x +2

3rd = x + 4

6(x + x + 4) = 24 + 11(x + 2)

6(2x + 4) = 24 + 11x + 22

12x + 24 = 46 + 11x

x + 24 = 46

x = 22

x + 2 = 24

x + 4 = 26

You might be interested in

Answer questions 25,26 now please please please help me

hoa [83]

The answers are very hard to see

7 0

3 years ago

Read 2 more answers

Help with a math problem

Nimfa-mama [501]

Answer:

For B the answer is 1 : 30. For A the graph is Y goes up 2 for every 60 on the X.

3 0

3 years ago

Read 2 more answers

5 divided by 978 with remainder

grigory [225]

Hi!

978 can't go into 5 at all, so the answer is 0 R: 978

Hope this helps! :)

-Peredhel

6 0

3 years ago

Read 2 more answers

Frorn first to last, when did the presidents serve?

navik [9.2K]

Step-by-step explanation:

you got this i believe in you

8 0

3 years ago

Which is the value of this expression when p=-2 and q=-1?

saul85 [17]

Answer:

D. 4

Step-by-step explanation:

$[(p^2) (q^{-3}) ]^{-2}.[(p)^{-3}(q)^5] ^{-2}\\\\=[(p^2) (q^{-3}) \times(p)^{-3}(q)^5 ]^{-2}\\\\=[(p^{2}) \times(p)^{-3} \times (q^{-3}) \times(q)^5 ]^{-2}\\\\=[(p^{2-3}) \times (q^{5-3}) ]^{-2}\\\\=[(p^{-1}) \times (q^{2}) ]^{-2}\\\\=(p^{-1\times (-2)}) \times (q^{2\times (-2) }) \\\\=p^{2}\times q^{-4} \\\\= \frac{p^2}{q^4}\\\\= \frac{(-2)^2}{(-1)^4}\\\\= \frac{4}{1}\\\\= 4$

8 0

3 years ago