(y-4)(y^2+4y+16)
y^3+4y^2+16y-4y^2-16y-64=y^3+4y^2+ay-4y^2-ay-64
Comparing the coefficients: a=16
Answer: The value of a in the polynomial is 16
Their is not a lot of information but I think it is 12*24=288
Answer:
128
Step-by-step explanation:
2 to the power of 7 means you multiply 2 seven times over
2*2*2*2*2*2*2 = 4*2*2*2*2*2= 8*2*2*2*2= 16*2*2*2= 32*2*2= 64*2 = 128
plz give brainliest
Answer:
y = 0
Step-by-step explanation:
It is always a good idea to look at the question and make some observations about it. Here, you might observe ...
- all of the bases are powers of 3: 243 = 3^5; 9 = 3^2
- y is a factor of every exponent
The latter observation is important, because it means that when y=0, every exponential expression has a value of 1. Hence y = 0 is a solution.
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To solve the equation, you can write it in terms of powers of 3.
(3^5)^(-y) = (3^-5)^(3y)·(3^2)^(-2y)
3^(-5y) = 3^(-15y)·3^(-4y)
3^(-5y) = 3^(-19y)
-5y = -19y . . . . . . . . equating exponents; equivalent to taking log base 3
14y = 0 . . . . . . . . . . add 19y
y = 0 . . . . . . . . . . . one solution
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The rules of exponents we used are ...
(a^b)(a^c) = a^(b+c)
(a^b)^c = a^(bc)
1/a^b = a^-b
One of the product rules for exponents was used.
Or, you can consider that the definition of an exponent and the associative and commutative properties of multiplication were used.
