Mike age is 3 less then twice the age of his brother nuke.
M = Mike's age now
N = Nuke's age now
M = 2N - 3

Ten years ago, four times nuke's age was one more mike's age.
M-10 = Mike's age 10 years ago
N-10 = Muke's age 10 years ago
4(N - 10) = 1 + M -10

what are their present ages

find M and N

substitute M = 2N - 3 into 4(N - 10) = 1 + M -10:

4(N - 10) = 1 + [2N - 3] - 10
4N - 40 = 2N - 12
2N = 28
N = 14

M = 2(14) - 3 = 28 - 3 = 25

Mike is 25 and Nuke is 14.

check:
4(14 - 10) = 4(4) = 16 = 26 -10 = 1 + 25 - 10

4 0

3 years ago

Simplyfy 8x^4y^5 times 10xy^3

Dennis_Churaev [7]

Answer:

$80x^5y^8$

Step-by-step explanation:

$8x^4y^5\times 10xy^3=\\\\(8\times 10) \times (x^4\times x)\times (y^5\times y^3)=\\\\80x^5y^8$

Hope this helps!

4 0

3 years ago

Read 2 more answers