Algebra II

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Questions 1 - 10

Simplify:

$\frac{6^{12}}{6^{18}}$

$6^{-6}$

$6^{30}$

$6^{15}$

$-6^{6}$

$6^{\frac{2}{3}}$

Explanation

When dividing exponents with the same base, we just subtract the exponents.

$\frac{6^{12}}{6^{18}}=6^{12-18}=6^{-6}$

$\frac{2-x}{x^2}+\frac{4}{x}$

$\frac{3x+2}{x^2}$

$-\frac{3x+2}{x^2}$

$\frac{3x-2}{x^2}$

$\frac{3x+2}{x}$

$\frac{3}{x^2}$

Explanation

To combine these rational expressions, first find the common denominator. In this case, it is . Then, offset the second equation so that you get the correct denominator: $\frac{4}{x}=\frac{4x}{x^2}$ . Then, combine the numerators: . Put your numerator over the denominator for your answer: $\frac{3x+2}{x^2}$ .

Simplify:

$10^{-5}$

$\frac{1}{10^5}$

$-\frac{1}{10^5}$

Explanation

When an exponent is negative, we express as such:

$x^{-a}=\frac{1}{x^a}$

is the positive exponent, and is the base.

$10^{-5}=\frac{1}{10^5}= \frac{1}{100000}$ .

Simplify.

$\frac{x^5}{x^4}$

$x^{1.25}$

$x^{20}$

$x^{-1}$

Explanation

When dividing exponents, we must first check to see if the factors have the same base. If they have the same bases, then we can subtract the exponents.

$\frac{x^5}{x^4}=x^{5-4}$

$x^{5-4}=x^{1}=x$

$\frac{2-x}{x^2}+\frac{4}{x}$

$\frac{3x+2}{x^2}$

$-\frac{3x+2}{x^2}$

$\frac{3x-2}{x^2}$

$\frac{3x+2}{x}$

$\frac{3}{x^2}$

Explanation

Simplify: $\frac{5}{x}-\frac{2x}{x+7}$

$-\frac{2x^2-5x-35}{x^2+7x}$

$-\frac{2x^2+5x-35}{x^2+7x}$

$-\frac{2x^2-5x-35}{x+7}$

$\frac{2x^2-5x-35}{x+7}$

$\frac{2x-5}{x-7}$

Explanation

In order to add the numerators, we will need the least common denominator.

Multiply the denominators together.

Convert both fractions by multiplying both the top and bottom by what was multiplied to get the denominator. Rewrite the fractions and combine as one single fraction.

$\frac{5(x+7)}{x(x+7)}-\frac{2x(x)}{(x)(x+7)} = \frac{5x+35-2x^2}{x^2+7x}$

Re-order the terms.

$\frac{-2x^2+5x+35}{x^2+7x}$

Pull out a common factor of negative one on the numerator.

$\frac{-1(2x^2-5x-35)}{x^2+7x} = -\frac{2x^2-5x-35}{x^2+7x}$

The answer is: $-\frac{2x^2-5x-35}{x^2+7x}$

Which value of makes the following expression undefined?

$\frac{x+1}{x-1}$

$\text{All real numbers}$

$\text{Cannot be determined}$

Explanation

A rational expression is undefined when the denominator is zero.

The denominator is zero when .

Evaluate $2^{\frac{1}2{}}$

$\sqrt{2}$

$\frac{1}{4}$

$\frac{\sqrt{2}}2{}$

Explanation

When dealing with fractional exponents, remember this form:

$\sqrt[a]{x^n}$ is the index of the radical which is also the denominator of the fraction, represents the base of the exponent, and is the power the base is raised to. That value is the numerator of the fraction.

$2^{\frac{1}{2}}=\sqrt[2]{2^1}=\sqrt{2}$

Solve the absolute value equation: $\left | 9x-9\right | = -1$

$\textup{No solution.}$

$\frac{8}{9},\frac{10}{9}$

$-\frac{8}{9},-\frac{10}{9}$

$-\frac{8}{9},\frac{10}{9}$

$\frac{8}{9},-\frac{10}{9}$

Explanation

Recall that the absolute value sign will convert any value to a positive sign. There will be no occurrences of that will evaluate into a negative one as a final solution.

There are no solutions for this equation.

The answer is: $\textup{No solution.}$

What is $4\cdot (7+2)$ ?

Explanation

Remember PEMDAS, the acronym that stands for Parentheses, Exponents, Multiplication, Division, Addition, and Subtraction. This is the order of operations we need to adhere to when doing any arithmetic.

The parentheses go first so do what's inside the parentheses.

The sum of and is .