Step-by-step explanation:
In order to be like terms, the variables and exponents must be the same.
3a can be combined with 14a and 4a.
4b can be combined with 3b and 16b.
a² cannot be combined with any of the terms.
(2^8 *3^-5* 6^0)^-2 * ((3^-2)/(2^3))^4 * 2^28
anything to the 0 power is 1
(2^8 *3^-5* 1)^-2 * ((3^-2)/(2^3))^4 * 2^28
using the power of power property to take the power inside
(2^(8*-2) *3^(-5* -2) * (3^-2*4)/(2^3*4) * 2^28
simplify
2^ -16 * 3^10 * 3^-8 /2*12 * 2^28
get rid of the division by making the exponent negative
2^-16 * 3^10 * 3^-8 *2*-12 * 2^28
combine exponents with like bases
2^(-16-12+28) * 3^(10-8)
2^(0) *3^2
anything to the 0 power is 1
1*9
9
Answer:
The Answer is C.
Step-by-step explanation:
Answer:
It is <em>a</em> solution, one of an infinite number of solutions.
Step-by-step explanation:
You can check to see if the given point is in the solution set:
4(-4) +5(4) = -16 +20 = 4 > -6 . . . . yes
-2(-4) +7(4) = 8 +28 = 36 > 20 . . . yes
The offered point satisfies both inequalities, so is in the solution set. It is not <em>the</em> solution, but is one of an infinite number of solutions.
