Mathematical Relationships and Basic Graphs - Math

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Question

$2x + \left | x - 3 \right | = -12$

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Answer

Notice that the equation $2x + \left | x - 3 \right | = -12$ has an term both inside and outside the absolute value expression.

Since the absolute value expression will always produce a positive number and the right side of the equation is negative, a negative number must be added to the result of the absolute value expression to satisfy the equation. Therefore the term outside of the absolute value expression (in this case ) must be negative (meaning must be negative).

Since will be a negative number, the expression within the absolute value will also be negative (before the absolute value is taken). It is thus possible to convert the original equation into an equation that treats the absolute value as a parenthetical expression that will be multiplied by , since any negative value becomes its opposite when taking the absolute value.

$2x + \left | x - 3 \right | = -12$

$\rightarrow 2x+ (x-3)(-1) = -12$

Simplifying and solving this equation for gives the answer:

$\rightarrow 2x -x+3 = -12$

$\rightarrow x + 3 = -12$

$\rightarrow x = -15$

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Question

$2x + \left | x - 3 \right | = -12$

Tap to reveal answer

Answer

Notice that the equation $2x + \left | x - 3 \right | = -12$ has an term both inside and outside the absolute value expression.

Since the absolute value expression will always produce a positive number and the right side of the equation is negative, a negative number must be added to the result of the absolute value expression to satisfy the equation. Therefore the term outside of the absolute value expression (in this case ) must be negative (meaning must be negative).

Since will be a negative number, the expression within the absolute value will also be negative (before the absolute value is taken). It is thus possible to convert the original equation into an equation that treats the absolute value as a parenthetical expression that will be multiplied by , since any negative value becomes its opposite when taking the absolute value.

$2x + \left | x - 3 \right | = -12$

$\rightarrow 2x+ (x-3)(-1) = -12$

Simplifying and solving this equation for gives the answer:

$\rightarrow 2x -x+3 = -12$

$\rightarrow x + 3 = -12$

$\rightarrow x = -15$

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Question

Solve for :

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Answer

To solve for in the equation

Factor out of the expression on the left of the equation:

$\rightarrow 2(x^2 - 16) =0$

Use the "difference of squares" technique to factor the parenthetical term on the left side of the equation.

$\rightarrow 2(x+4)(x-4)=0$

Any variable that causes any one of the parenthetical terms to become will be a valid solution for the equation. becomes when is , and becomes when is , so the solutions are and .

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Question

Solve for :

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Answer

Pull an out of the left side of the equation.

$\rightarrow x(x^2 - 9) =0$

Use the difference of squares technique to factor the expression in parentheses.

$\rightarrow x(x+3)(x-3) =0$

Any number that causes one of the terms , , or to equal is a solution to the equation. These are , , and , respectively.

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Question

Solve for :

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Answer

To solve for in the equation

Factor out of the expression on the left of the equation:

$\rightarrow 2(x^2 - 16) =0$

Use the "difference of squares" technique to factor the parenthetical term on the left side of the equation.

$\rightarrow 2(x+4)(x-4)=0$

Any variable that causes any one of the parenthetical terms to become will be a valid solution for the equation. becomes when is , and becomes when is , so the solutions are and .

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Question

Solve for :

Tap to reveal answer

Answer

Pull an out of the left side of the equation.

$\rightarrow x(x^2 - 9) =0$

Use the difference of squares technique to factor the expression in parentheses.

$\rightarrow x(x+3)(x-3) =0$

Any number that causes one of the terms , , or to equal is a solution to the equation. These are , , and , respectively.

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Question

Solve for :

Tap to reveal answer

Answer

To solve for in the equation

Factor out of the expression on the left of the equation:

$\rightarrow 2(x^2 - 16) =0$

Use the "difference of squares" technique to factor the parenthetical term on the left side of the equation.

$\rightarrow 2(x+4)(x-4)=0$

Any variable that causes any one of the parenthetical terms to become will be a valid solution for the equation. becomes when is , and becomes when is , so the solutions are and .

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Question

Solve for :

Tap to reveal answer

Answer

Pull an out of the left side of the equation.

$\rightarrow x(x^2 - 9) =0$

Use the difference of squares technique to factor the expression in parentheses.

$\rightarrow x(x+3)(x-3) =0$

Any number that causes one of the terms , , or to equal is a solution to the equation. These are , , and , respectively.

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Question

Solve for :

Tap to reveal answer

Answer

To solve for in the equation

Factor out of the expression on the left of the equation:

$\rightarrow 2(x^2 - 16) =0$

Use the "difference of squares" technique to factor the parenthetical term on the left side of the equation.

$\rightarrow 2(x+4)(x-4)=0$

Any variable that causes any one of the parenthetical terms to become will be a valid solution for the equation. becomes when is , and becomes when is , so the solutions are and .

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Question

Solve for :

Tap to reveal answer

Answer

Pull an out of the left side of the equation.

$\rightarrow x(x^2 - 9) =0$

Use the difference of squares technique to factor the expression in parentheses.

$\rightarrow x(x+3)(x-3) =0$

Any number that causes one of the terms , , or to equal is a solution to the equation. These are , , and , respectively.

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Question

$2x + \left | x - 3 \right | = -12$

Tap to reveal answer

Answer

Notice that the equation $2x + \left | x - 3 \right | = -12$ has an term both inside and outside the absolute value expression.

Since the absolute value expression will always produce a positive number and the right side of the equation is negative, a negative number must be added to the result of the absolute value expression to satisfy the equation. Therefore the term outside of the absolute value expression (in this case ) must be negative (meaning must be negative).

Since will be a negative number, the expression within the absolute value will also be negative (before the absolute value is taken). It is thus possible to convert the original equation into an equation that treats the absolute value as a parenthetical expression that will be multiplied by , since any negative value becomes its opposite when taking the absolute value.

$2x + \left | x - 3 \right | = -12$

$\rightarrow 2x+ (x-3)(-1) = -12$

Simplifying and solving this equation for gives the answer:

$\rightarrow 2x -x+3 = -12$

$\rightarrow x + 3 = -12$

$\rightarrow x = -15$

← Didn't Know|Knew It →

Question

$2x + \left | x - 3 \right | = -12$

Tap to reveal answer

Answer

Notice that the equation $2x + \left | x - 3 \right | = -12$ has an term both inside and outside the absolute value expression.

Since the absolute value expression will always produce a positive number and the right side of the equation is negative, a negative number must be added to the result of the absolute value expression to satisfy the equation. Therefore the term outside of the absolute value expression (in this case ) must be negative (meaning must be negative).

Since will be a negative number, the expression within the absolute value will also be negative (before the absolute value is taken). It is thus possible to convert the original equation into an equation that treats the absolute value as a parenthetical expression that will be multiplied by , since any negative value becomes its opposite when taking the absolute value.

$2x + \left | x - 3 \right | = -12$

$\rightarrow 2x+ (x-3)(-1) = -12$

Simplifying and solving this equation for gives the answer:

$\rightarrow 2x -x+3 = -12$

$\rightarrow x + 3 = -12$

$\rightarrow x = -15$

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Question

Solve for :

$\sqrt{x^2+12x+36} =4$

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Answer

To solve for in the equation $\sqrt{x^2+12x+36} =4$

Square both sides of the equation

$\rightarrow (\sqrt{x^2+12x+36})^2 =(4)^2$

$\rightarrow x^2+12x+36 =16$

Set the equation equal to by subtracting the constant from both sides of the equation.

$\rightarrow (x^2+12x+36) - 16 =(16) -16$

$\rightarrow x^2+12x+20 =0$

Factor to find the zeros:

$\rightarrow (x+10)(x+2) = 0$

This gives the solutions

$x = \left \{ -10, -2\right \}$ .

Verify that these work in the original equation by substituting them in for . This is especially important to do in equations involving radicals to ensure no imaginary numbers (square roots of negative numbers) are created.

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Question

Solve for :

$\sqrt{x^2+12x+36} =4$

Tap to reveal answer

Answer

To solve for in the equation $\sqrt{x^2+12x+36} =4$

Square both sides of the equation

$\rightarrow (\sqrt{x^2+12x+36})^2 =(4)^2$

$\rightarrow x^2+12x+36 =16$

Set the equation equal to by subtracting the constant from both sides of the equation.

$\rightarrow (x^2+12x+36) - 16 =(16) -16$

$\rightarrow x^2+12x+20 =0$

Factor to find the zeros:

$\rightarrow (x+10)(x+2) = 0$

This gives the solutions

$x = \left \{ -10, -2\right \}$ .

Verify that these work in the original equation by substituting them in for . This is especially important to do in equations involving radicals to ensure no imaginary numbers (square roots of negative numbers) are created.

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Question

Solve for :

$\sqrt{x^2+12x+36} =4$

Tap to reveal answer

Answer

To solve for in the equation $\sqrt{x^2+12x+36} =4$

Square both sides of the equation

$\rightarrow (\sqrt{x^2+12x+36})^2 =(4)^2$

$\rightarrow x^2+12x+36 =16$

Set the equation equal to by subtracting the constant from both sides of the equation.

$\rightarrow (x^2+12x+36) - 16 =(16) -16$

$\rightarrow x^2+12x+20 =0$

Factor to find the zeros:

$\rightarrow (x+10)(x+2) = 0$

This gives the solutions

$x = \left \{ -10, -2\right \}$ .

Verify that these work in the original equation by substituting them in for . This is especially important to do in equations involving radicals to ensure no imaginary numbers (square roots of negative numbers) are created.

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Question

Solve for :

$\sqrt{x^2+12x+36} =4$

Tap to reveal answer

Answer

To solve for in the equation $\sqrt{x^2+12x+36} =4$

Square both sides of the equation

$\rightarrow (\sqrt{x^2+12x+36})^2 =(4)^2$

$\rightarrow x^2+12x+36 =16$

Set the equation equal to by subtracting the constant from both sides of the equation.

$\rightarrow (x^2+12x+36) - 16 =(16) -16$

$\rightarrow x^2+12x+20 =0$

Factor to find the zeros:

$\rightarrow (x+10)(x+2) = 0$

This gives the solutions

$x = \left \{ -10, -2\right \}$ .

Verify that these work in the original equation by substituting them in for . This is especially important to do in equations involving radicals to ensure no imaginary numbers (square roots of negative numbers) are created.

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Question

What is the value of ?

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Answer

! is the symbol for factorial, which means the product of the whole numbers less than the given number.

Thus, .

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Question

What is the value of ?

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Answer

By definition, $4! = 4 \cdot 3 \cdot 2 \cdot 1= 24$ .

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Question

List the first 4 terms of an arithmetic sequence with a first term of 3, and a common difference of 5.

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Answer

An arithmetic sequence is one in which the common difference is added to one term to get the next term. Thus, if the first term is 3, we add 5 to get the second term, and continue in this manner.

Thus, the first four terms are:

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Question

Which of the following is a geometric sequence?

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Answer

A geometric sequence is one in which the next term is found by mutlplying the previous term by a particular constant. Thus, we look for an implicit definition which involves multiplication of the previous term. The only possibility is:

$t_{0}=4$

$t_{n}=t_{n-1}\times 3$

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