Answer:
Step-by-step explanation:
Before you add 2 fractions, the denominator needs to be the same. If they are the same, which they are, then add the 2 numerators.
1 plus 2 equals 3.
So the answer is 3 over 5.
Given : f(x) = 3x + 1
We get f(2) by substituting x = 2 in the given Function
⇒ f(2) = 3(2) + 1
⇒ f(2) = 6 + 1 = 7
Option C is the Answer
Answer: 0.375
Step-by-step explanation:
Answer: D
4x^4 + 3x^3 + 2x^2 - 11x -5
Step-by-step explanantion:
Flip and make the 4x-5 in the front and then multiply all terms by 4x and then all terms by -5 and then add both answers together.
Answer:
use logarithms
Step-by-step explanation:
Taking the logarithm of an expression with a variable in the exponent makes the exponent become a coefficient of the logarithm of the base.
__
You will note that this approach works well enough for ...
a^(x+3) = b^(x-6) . . . . . . . . . . . variables in the exponents
(x+3)log(a) = (x-6)log(b) . . . . . a linear equation after taking logs
but doesn't do anything to help you solve ...
x +3 = b^(x -6)
There is no algebraic way to solve equations that are a mix of polynomial and exponential functions.
__
Some functions have been defined to help in certain situations. For example, the "product log" function (or its inverse) can be used to solve a certain class of equations with variables in the exponent. However, these functions and their use are not normally studied in algebra courses.
In any event, I find a graphing calculator to be an extremely useful tool for solving exponential equations.
