Mathematical Relationships and Basic Graphs

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Algebra 2 › Mathematical Relationships and Basic Graphs

Questions 1 - 10

Solve: $-2i^2-4i^{-4}+6i^6$

Explanation

Evaluate each term of the expression. Write out the values of the imaginary terms.

$i=\sqrt{-1}$

$i^2= i\cdot i = -1$

$i^4 = i^2\cdot i^2 = (-1)\cdot (-1) = 1$

$i^6 = i^4 \cdot i^2 = (1)\cdot (-1) = -1$

Replace the values of each.

$-4i^{-4} = \frac{-4}{i^4} = -\frac{4}{1} = -4$

Sum all the values.

$-2i^2-4i^{-4}+6i^6 = 2-4-6 = -8$

The answer is:

$What\ is\ the\ complex\ conjugate\ for\ 9-2i?$

$81-4i^{2}$

$81+4i^{2}$

Explanation

$The\ complex\ conjugate\ for\ a binomial\ with\ an\ imaginary\ number\ is$

$simply\ the\ original\ with\ the\ sign\ for\ the\ imaginary\ number\ changed.$

$If\ 9-2i\ were\ multiplied\ by\ 9+2i\ the\ i\ values\ would\ cancel\ producing\ a\ real\ number.$

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.

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

Evaluate $17^\frac{1}{2}$

$\sqrt{17}$

$\sqrt{8.5}$

Explanation

When dealing with fractional exponents, we rewrite as such:

$x^\frac{a}{b}=\sqrt[b]{x^a}$

in which is the index of the radical and is the exponent raising base .

$17^\frac{1}{2}=\sqrt{17}$

Given the sequence , what is the 7th term?

Explanation

The formula for geometric sequences is defined by:

$ar^{n-1}$

The term represents the first term, while is the common ratio. The term represents the terms.

Substitute the known values.

$3(3)^{n-1}$

To determine the seventh term, simply substitute into the expression.

$3(3)^{7-1}$

The answer is:

State the domain of the function:

$y=\sqrt{4x+3}$

$x\geq-\frac{3}{4}$

$x\geq\frac{3}{4}$

$x\geq0$

$x\leq0$

$x\geq -\frac{4}{3}$

Explanation

Since the expression under the radical cannot be negative,

$4x+3\geq0$ .

Solve for x:

$x\geq-\frac{3}{4}$

This is the domain, or possible values, for the function.

Add the fractions: $\frac{5}{2}+\frac{9}{4}+\frac{13}{6}$

$\frac{83}{12}$

$\frac{37}{6}$

$\frac{23}{3}$

$\frac{73}{12}$

$\frac{41}{6}$

Explanation

Convert the fractions to a least common denominator in order to add the numerators. Write out the factors for each denominator to determine the LCD.

The LCD is 12.

$\frac{5}{2}+\frac{9}{4}+\frac{13}{6} =\frac{5(6)}{2(6)}+\frac{9(3)}{4(3)}+\frac{13(2)}{6(2)}$

Simplify the fractions.

$\frac{30}{12}+\frac{27}{12}+\frac{26}{12} = \frac{83}{12}$

The answer is: $\frac{83}{12}$

Simplify the following expression.

$\frac{25b^4}{5b^2}$

Explanation

Simplify the following expression

$\frac{25b^4}{5b^2}$

To simplify this, we need to subtract our exponents and divide our whole numbers.

When we do this, we get the following.

$\frac{25b^4}{5b^2}=\frac{25}{5}\frac{b^4}{b^2}=\frac{5\cdot 5b^2\cdot b^2}{5b^2}=5b^2$

Give the vertex of the graph of the function .

None of the other choices gives the correct response.

Explanation

Let

The graph of this basic absolute value function is a "V"-shaped graph with a vertex at the origin, or the point with coordinates . In terms of ,

or, alternatively written,

The graph of is the same as that of , after it shifts 10 units left ( ), it flips vertically (negative symbol), and it shifts up 10 units (the second ). The flip does not affect the position of the vertex, but the shifts do; the vertex of the graph of is at .

Solve for .

Explanation

When we add exponents, we try to factor to see if we can simplify it. Let's factor . We get $2^x(1+1)=2^x*2^1=2^{x+1}$ . Remember to apply the rule of multiplying exponents which is to add the exponents and keeping the base the same.

$2^{x+1}=2^6$ With the same base, we can rewrite as .

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