If
, then
![\sqrt[2]{y}=\sqrt[2]{x^2}=x](https://tex.z-dn.net/?f=%5Csqrt%5B2%5D%7By%7D%3D%5Csqrt%5B2%5D%7Bx%5E2%7D%3Dx)
if
, then
![\sqrt[3]{y}=\sqrt[3]{x^3}=x](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7By%7D%3D%5Csqrt%5B3%5D%7Bx%5E3%7D%3Dx)
ok
so
![\sqrt[2]{64}=\sqrt[2]{8^2}=8](https://tex.z-dn.net/?f=%5Csqrt%5B2%5D%7B64%7D%3D%5Csqrt%5B2%5D%7B8%5E2%7D%3D8)
![\sqrt[3]{64}=\sqrt[3]{4^3}=4](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B64%7D%3D%5Csqrt%5B3%5D%7B4%5E3%7D%3D4)
well, the cube root usually bigger
<em>The answer to your problem is:</em>
x = 1/6
Answer:
See below.
Step-by-step explanation:
Part 1.
x + 12 <---------- Quotient.
------------------
x - 4 ( x^2 + 8x + 16
x^2- 4x
------------
12x + 16
12x - 48
----------
64 <--------- Remainder.
Part 2
f(4) = (4)^2 +8(4) + 16 = 16 + 32 + 16
= 64 Which is the remainder we found in the long division.
Part 3.
As you see in Parts 1 and 2, the Remainder Theorem tells you what the remainder is without doing the long division. If the remainder is 0 this means that the binomial you is a factor of the polynomial.
Answer:
d = 2
Step-by-step explanation:
Generate the first few terms
a₁ = (2 × 1) - 4 = 2 - 4 = -2
a₂ = (2 × 2) - 4 = 4 - 4 = 0
a₃ = (2 × 3) - 4 = 6 - 4 = 2
a₄ = (2 × 4) - 4 = 8 - 4 = 4
d = 0 - (- 2) = 2 - 0 = 4 - 2 = 2
Answer:
1=0 2
Step-by-step explanation:
