Algebra II - Math

Card 0 of 9272

Solve for .

$4\left | x\right |+93=145$

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$4\left | x\right |+93=145$ Subtract on both sides.

$4\left | x\right |=52$ Divide on both sides.

$\left | x\right |=13$ Since it's absolute value, we need to accept both positive and negative answers.

$x=\pm13$

$\small $$\frac{x^2$-5x-6}{x+1}$$

For all values $x\neq -1$ , which of the following is equivalent to the expression above?

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First, factor the numerator. We need factors that multiply to $\small -6$ and add to $\small -5$ .

$\small 1*-6=-6\ $\text{and}$\ 1+(-6)=-5$

$\small $x^2$-5x-6=(x-6)(x+1)$

We can plug the factored terms into the original expression.

$\small $$\frac{x^2$-5x-6}{x+1}$=\frac{(x-6)(x+1)}{(x+1)}$$

Note that $\small (x+1)$ appears in both the numerator and the denominator. This allows us to cancel the terms.

$$\frac{(x-6)(x+1)}{(x+1)}$=(x-6)$

This is our final answer.

Simplify the expression.

$\small $$\frac{(x^2y$^$2)(x^3y$^$4z)}{x^4y$^$3z^4$}$$

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$\small $$\frac{(x^2y$^$2)(x^3y$^$4z)}{x^4y$^$3z^4$}$$

Because we are only multiplying terms in the numerator, we can disregard the parentheses.

$\small $$\frac{(x^2y$^$2)(x^3y$^$4z)}{x^4y$^$3z^4$$}$=\frac{x^2y$^$2x^3y$^$4z}{x^4y$^$3z^4$}$$

To combine like terms in the numerator, we add their exponents.

$\small $$\frac{x^5y$^$6z}{x^4y$^$3z^4$}$$

To combine like terms between the numerator and denominator, subtract the denominator exponent from the numerator exponent.

Remember that any negative exponents stay in the denominator.

Give the value of that makes the polynomial the square of a linear binomial.

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A quadratic trinomial is a perfect square if and only if takes the form

for some values of and .

, so

and .

For to be a perfect square, it must hold that

$Nx = 2AB = 2 \cdot 3x \cdot 10 = 60x$ ,

so . This is the correct choice.

Exponentiate:

$\left ( 9R - 10\right $)^{3}$$

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The difference of two terms can be cubed using the pattern

Where :

How many of the following are prime factors of ?

I)

II)

III)

IV)

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Factor all the way to its prime factorization.

can be factored as the difference of two perfect square terms as follows:

$=\left ( $9y^{2}$ \right $)^{2}$ - $1^{2}$$

$=\left (9 $y^{2}$ + 1 \right )\left ( $9y^{2}$ - 1 \right )$

is a factor, and, as the sum of squares, it is a prime. is also a factor, but it is not a prime factor - it can be factored as the difference of two perfect square terms. We continue:

$\left (9 $y^{2}$ + 1 \right )\left ( $9y^{2}$ - 1 \right )$

$=\left ( $9y^{2}$ + 1 \right )\left [ \left (3y \right ) ^{2} - $1^{2}$ \right ]$

$=\left ( $9y^{2}$ + 1 \right )\left (3y+1 \right )\left (3 y-1\right )$

Therefore, of the given four choices, only is not a factor, so the correct response is three.

Factor:

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$=\left (3Q \right $)^{3}$- 3 \cdot $90Q^{2}$+3\cdot 300Q $-10^{3}$$

$=\left (3Q \right $)^{3}$- 3 \cdot $9Q^{2}$\cdot 10+3\cdot 3Q \cdot $10^{2}$ $-10^{3}$$

$=\left (3Q \right $)^{3}$- 3 \cdot \left (3Q \right $)^{2}$\cdot 10+3\cdot 3Q \cdot $10^{2}$ $-10^{3}$$

This can be factored out as the cube of a difference, where :

Therefore,

$=\left (3Q \right $)^{3}$- 3 \cdot \left (3Q \right $)^{2}$\cdot 10+3\cdot 3Q \cdot $10^{2}$ $-10^{3}$$

$= (3Q - $10)^{3}$$

Simplify:

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To simplify, we begin by simplifying the numerator. When muliplying like bases with different exponents, their exponents are added.

For x: $\small 4+3=7$

For y: $\small 2+1=3$

For z: $\small 3+1=4$

The numerator is now .

When dividing like bases, their exponents are subtracted.

For x: $\small 7-4=3$

For y: $\small 3-(-3)=6$

For z: $\small 4-3=1$

Thus, our answer is .

Which of the following linear binomials is a factor of the polynomial ?

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By the factor theorem, a polynomial is divisible by the linear binomial if and only if . We can use this fact to test each of the binomials by evaluating the dividend for the appropriate value of .

: Evaluate the polynomial at :

: Evaluate the polynomial at :

: Evaluate the polynomial at :

: Evaluate the polynomial at :

: Evaluate the polynomial at :

The dividend assumes the value of 0 at , so of the choices given, is the factor.

What is increased by 40%?

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A number increased by 40% is equivalent to 100% of the number plus 40% of the number. This is taking 140% of the number, or, equivalently, multiplying it by 1.4.

Therefore, increased by 40% is 1.4 times this, or

$1.4 \left (3t+ 7 \right ) = 1.4 \cdot 3t + 1.4 \cdot 7 = 4.2 t + 9.8$

Simplify the expression:

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To solve this problem, we first need to factor the numerator. We are looking for two numbers that multiply to equal -8 and sum to equal 2.

Now, we can write out our expression in fraction form.

$\frac{(x+4)(x-2)}{x+4}$

Since we have the like term in the numerator and denominator, we can cancel them out of our expression.

$$\frac{(x+4)(x-2)}{x+4}$= x-2$

Thus, our answer is .

Which of the following is a prime factor of ?

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is the difference of two squares:

As such, it can be factored as follows:

The first factor is the sum of cubes and the second is the difference of cubes; each can be factored further:

Therefore,

Of the choices, appears in the prime factorization and is therefore the correct choice.

Decrease by 20%. Which of the following will this be equal to?

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A number decreased by 20% is equivalent to 100% of the number minus 20% of the number. This is taking 80% of the number, or, equivalently, multiplying it by 0.8.

Therefore, decreased by 20% is 0.8 times this, or

$0.8 \left (9x + 4y \right ) = 0.8 \cdot 9x + 0.8 \cdot 4y = 7.2x + 3.2y$

Divide:

$\left ( 12 x ^{4} + $16x^{3}$+ $64x^{2}$+8 x \right ) \div $8x^{2}$$

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Divide termwise:

$\left ( 12 x ^{4} + $16x^{3}$+ $64x^{2}$+8 x \right ) \div $8x^{2}$$

$=\frac{ 12 x ^{4}$ + $16x^{3}$+ $64x^{2}$+8 x }{ $8x^{2}$}$

$=\frac{ 12 x ^{4}$ }{ $8x^{2}$} + $\frac{ $16x^{3}$$ }{ $8x^{2}$} + $\frac{ $64x^{2}$$ }{ $8x^{2}$} +\frac{ 8 x }{ $8x^{2}$$}$

$=\frac{ 12 }{ 8 }$x ^{4-2} + $\frac{ 16 }{ $8}$x^{3-2}$ + $\frac{ 64 }{ 8 }$ +\frac{ 8 }{ 8}$ \cdot $$\frac{1}{x^{2-1}$$}$

$=\frac{ 3 }{ 2 }$x ^{2} + 2x + 8 + $\frac{1}{x}$$

Decrease by 30%. Which of the following will this be equal to?

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A number decreased by 30% is equivalent to 100% of the number minus 30% of the number. This is taking 70% of the number, or, equivalently, multiplying it by 0.7.

Therefore, decreased by 30% is 0.7 times this, or

$0.7 \left (6y + 17 \right ) = 0.7 \cdot 6y +0.7 \cdot 17 = 4.2y + 11.9$

The polynomial is divisible by the linear binomial . Evaluate .

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By the factor theorem, a polynomial is divisible by the linear binomial if and only if . Therefore, we want the value of that makes the polynomial equal to 0 when evaluated at .

$1,296- 27 \cdot 36+ 30- A = 0$

Factor:

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can be rewritten as and is therefore the sum of two cubes. As such, it can be factored using the pattern

where .

$= $(5T+9)[(5T)^{2}$-5T \cdot 9 $+9^{2}$]$

$= $(5T+9)(25T^{2}$-45T +81)$

Exponentiate:

$\left $(x^{2}$ + x + 7 \right $)^{2}$$

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$\left $(x^{2}$ + x + 7 \right $)^{2}$ = \left $(x^{2}$ + x + 7 \right )\left $(x^{2}$ + x + 7 \right )$

Vertical multiplication is perhaps the easiest way to multiply trinomials.

$x ^{3 } + $x^{2}$ + 7x$

$$\underline{x ^{4 }$ + $x^{3 }$ + $7x^{2}$ ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; ; }$

Solve for .

$$\frac{x}{8}$-5=-2$

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To isolate the variable in the equation, perform the opposite operation to move all constants on one side of the equation and leaving the variable on the other side of the equation.

$$\frac{x}{8}$-5=-2$

Add on both sides. Since is greater than and is positive, our answer is positive. We treat as a normal subtraction.

$$\frac{x}{8}$=3$

Multiply on both sides.

Factor completely:

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The grouping technique works here:

$=\left $(t^{3}$ $+4t^{2}$ \right )-\left (9t +36 \right )$

$=\left (t ^{2} \cdot t $+t^{2}$ \cdot 4} \right )-\left (9 \cdot t +9 \cdot 4 \right )$

$=t ^{2} \left (t + 4} \right )-9 \left ( t + 4 \right )$

$=\left (t ^{2}-9 \right ) \left (t + 4} \right )$

The first factor is the difference of squares and can be factored further accordingly:

$\left (t ^{2}-9 \right ) \left (t + 4} \right )$

$=\left (t ^{2}- $3^{2}$ \right ) \left (t + 4} \right )$

$=\left (t -3 \right ) \left (t +3 \right ) \left (t + 4} \right )$