6a
10a-37=6a+51 -6a -6a subtract 6a from both sides and the 6a's cancel 4a-37=51 +37 +37 4a=88 /4 /4 a=22
22
if the digits are A and B, then we are given:A + B = 6A * B = (10A + B) / 3∴ B = 6 - A∴ A * (6 - A) = (10A + [6 - A]) / 3∴ 6A - A2 = (9A + 6) / 3∴ 6A - A2 = 3A + 2∴ 6A - A2 - 3A - 2 = 0∴ A2 - 3A + 2 = 0∴(A - 2)(A - 1) = 0∴A ∈ {1, 2}A + B = 6∴ B ∈ {5, 4}∴ The numbers 15 or 24 both meet these conditions.
36a2 - 60a + 25 = 36a2 - 30a - 30a + 25 = 6a(6a - 5) - 5(6a - 5) = (6a - 5)(6a - 5) = (6a - 5)2
no
6a-10a=16 -4a=16 a=-4
6a
No, the voltage rating should not be exceeded. However, you could use a 250v 10a fuse for a 125v 10a fuse.
It is 6a.
10a-37=6a+51 -6a -6a subtract 6a from both sides and the 6a's cancel 4a-37=51 +37 +37 4a=88 /4 /4 a=22
no
22
to solve this expression: 10a-4(a+2) -> 10a-4(a)-4(2) -> 10a-4a-8 -> 6a-8
yes
10a - 10 b
8a + 5g