keeping in mind that when the logarithm base is omitted, the base 10 is assumed.
![\textit{exponential form of a logarithm} \\\\ \log_a(b)=y \qquad \implies \qquad a^y= b \\\\[-0.35em] ~\dotfill\\\\ \log(x)=2\implies \log_{10}(x)=2\implies 10^2=x\implies 100=x](https://tex.z-dn.net/?f=%5Ctextit%7Bexponential%20form%20of%20a%20logarithm%7D%20%5C%5C%5C%5C%20%5Clog_a%28b%29%3Dy%20%5Cqquad%20%5Cimplies%20%5Cqquad%20a%5Ey%3D%20b%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Clog%28x%29%3D2%5Cimplies%20%5Clog_%7B10%7D%28x%29%3D2%5Cimplies%2010%5E2%3Dx%5Cimplies%20100%3Dx)
Answer:
B(6 , -11)
Step-by-step explanation:
(x-2 , y+3) = (4 , -8)
Compare the x & y co ordinates
x - 2 = 4 ; y + 3 = -8
x = 4 +2 ; y = -8 - 3
x = 6 ; y = -11
B(6 , -11)
Equivalent Fractions for 9/25:
9/25, 18/50, 27/75, 36/100, 45/125, 54/150, 63/175, 72/200, 81/225, 90/250, and so on ...
Question:
Factor 
(8 - x)(8 - x)
(8 + x) (8 - x)
(X + 8)(x-8)
Answer:
Option B:
is the factor
Explanation:
Given that the expression is 
We need to find the factor of the given expression.
Let us rewrite the expression as 
Since the expression is of the form
, let us use the identity 
where a = 8 and b = x
Substituting the values in the identity, then the expression can be written as

Thus, the factor of the expression is 
Hence, Option B is the correct answer.
Answer:
poop
Step-by-step explanation:
poop