MrBillDoesMath!
Answer to #4: 81/256 * s^8 * t^ 12
Comments:
(7x^3) ^ (1/2) = 7 ^ (1/2) * x^(3/2) where ^(1/2) means the square root of a quantity. The answer written (7x^3) is NOT correct.
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(1) (27s^7t^11)^ (4/3)
= 27^(4/3) * (s^7)^(4/3) * (t^11)^ (4/3)
As 27 = 3^3, 27 ^(4/3) = 3^4 = 81
(2) (-64st^2)^ (4/3) = (-64)^(4/3) * (s^4/3) * t(^8/3)
As 64 = (-4)^3, (-64)^(4/3) = (-4)^4 = +256
So (1)/(2) =
81 * s^(28/3)* t^(44/3)
------------------------------- =
256 s^(4/3) * t^((8/3)
81/256 * s ^ (28/3 - 4/3) * t^(44/3 - 8/3) =
81/256 * s^(24/3) * t (36/3) =
81/256 * s^8 * t^ 12
MrB
The answer 0.6 i apologize if it is incorrect!
In order to use the elimination method, we have to multiply the equation by some number, so that one of the variable has the same coefficient.
For example, multiplying the first equation by 3 and the second by 2 gives the following, equivalent system:
Now, we can subtract the two equations, and we will cancel (eliminate) the x variable:
Now that y is known, plug it into one of the equations: for example, if we use the first one we get
Answer:
11.5
Step-by-step explanation:
1.50 x 5 = 7.50
0.74 + 3.25 = 4
7.50 + 4 = 11.5
Answer:
3(x+7) = 3x +21
Step-by-step explanation:
