Algebra - GED Math

Card 0 of 2050

Question

Sixty-four coins, all dimes and nickels, total $5.15. How many of the coins are dimes?

Answer

Let be the number of dimes. Then there are nickels.

An equation can be set up and solved for for the amount of money:

$0.10x + 0.05 \cdot 64-0.05 \cdot x = 5.15$

$0.05 x \div 0.05= 1.95 \div 0.05$

, the number of dimes.

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Question

Sixty-four coins, all dimes and quarters, total $8.95. How many quarters are there?

Answer

Let be the number of quarters. Then there are dimes.

An equation can be set up and solved for for the amount of money in dollars:

$0.25x + 0.10\cdot 64- 0.10\cdot x = 8.95$

$0.15x \div 0.15 = 2.55 \div 0.15$

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Question

Marge and William are running away from each other in opposite directions. Marge is running at a rate of $1.2;\frac{m}{s}$ , while William is running at a rate of $0.8;\frac{m}{s}$ . In how many minutes will they be $12;\textup{km}$ from each other?

Answer

Every second, you know that Marge and William will become a total of $1.2,\frac{m}{s}+0.8,\frac{m}{s}$ or $2.0,\frac{m}{s}$ . Now, you can use the simple work formula for distance:

or, for our data,

(Remember kilometers is $12\cdot1000=12000$ meters.)

Thus, solving for you get:

$\frac{12000}{2}=\frac{2T}{2}$

This is in seconds, though. You need minutes. To convert, you need to divide by :

$6000;\textup{s}\cdot\frac{1;\textup{min}}{60;\textup{s}}$

$\frac{6000;\textup{s}}{60;\textup{s}}=100;\textup{min}$

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Question

Timmy works at a fast food chain retail store five days a week, eight hours a day. Suppose it costs him $2.00 everyday to drive to and from work. He makes $10.00 per hour. How much will Timmy have at the end of the week, before applicable taxes?

Answer

Timmy makes ten dollars per hour for eight hours.

$$10 \cdot 8 = $80$

For five days: $$80 \cdot 5 = $400$

Timmy also will pay $$2 \cdot 5 = $10$ for the week to get to work and back.

Subtract his expense from his earnings for the week.

Timmy will have by the end of the week.

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Question

Multiply:

$\small -x(4x+2)=$

Answer

$\small -x(4x+2)=(-x)(4x)+(-x)(2)=-4x^2+(-2x)=-4x^2-2x$

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Question

Factor:

$\small x^2+5x+4$

Answer

$\small x^2+5x+4=(x+a)(x+b)$

where

$\small a+b=5\ and\ ab=4$

The numbers $\small 4$ and $\small 1$ fit those criteria. Therefore,

$\small x^2+5x+4=(x+1)(x+4)$

You can double check the answer using the FOIL method

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Question

Which of the following is a factor of the polynomial $x^{4}-13x^{2}+ 36$ ?

Answer

Perhaps the easiest way to identify the factor is to take advantage of the factor theorem, which states that is a factor of polynomial if and only if . We substitute 1, 2, 4, and 9 for in the polynomial to identify the factor.

:

$1^{4}-13 \cdot 1^{2}+ 36$

:

$2^{4}-13 \cdot 2^{2}+ 36$

$= 16- 13 \cdot 4 + 36$

:

$4^{4}-13 \cdot 4^{2}+ 36$

$= 256- 13 \cdot 16 + 36$

:

$9^{4}-13 \cdot 9^{2}+ 36$

$= 6,561- 13 \cdot 81 + 36$

Only makes the polynomial equal to 0, so among the choices, only is a factor.

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Question

Which of the following is a factor of the polynomial $x^{6} - 66x^{3} + 128$

Answer

Perhaps the easiest way to identify the factor is to take advantage of the factor theorem, which states that is a factor of polynomial if and only if . We substitute $\pm2$ and $\pm 4$ for in the polynomial to identify the factor.

:

$x^{6} - 66x^{3} + 128$

$2^{6} - 66 \cdot 2^{3} + 128$

$= 64 - 66 \cdot 8 + 128$

:

$x^{6} - 66x^{3} + 128$

$\left (-2 \right )^{6} - 66 \cdot \left (-2 \right )^{3} + 128$

$= 64 - 66 \cdot\left ( -8 \right )+ 128$

:

$x^{6} - 66x^{3} + 128$

$4^{6} - 66 \cdot 4^{3} + 128$

$= 4,096- 66 \cdot 64 + 128$

:

$x^{6} - 66x^{3} + 128$

$\left (-4 \right )^{6} - 66 \cdot \left (-4 \right )^{3} + 128$

$= 4,096- 66 \cdot\left ( -64 \right )+ 128$

Only makes the polynomial equal to 0, so of the four choices, only is a factor of the polynomial.

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Question

Which of the following is not a prime factor of $y^{4}- 81$ ?

Answer

Factor $y^{4}- 81$ all the way to its prime factorization.

$y^{4}- 81$ can be factored as the difference of two perfect square terms as follows:

$y^{4}- 81$

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

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

$y^{2} + 9$ is a factor, and, as the sum of squares, it is a prime. $y^{2} - 9$ 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:

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

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

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

Therefore, all of the given polynomials are factors of $y^{4}- 81$ , but $y^{2} - 9$ is the correct choice, as it is not a prime factor.

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Question

Which of the following is a prime factor of $x^{6} +1$ ?

Answer

$x^{6} +1$ is the sum of two cubes:

$x^{6} +1 =\left ( x^{2} \right ) ^{3}+1^{3}$

As such, it can be factored using the pattern

$A^{3}+ B^{3}= (A+B)(A^{2}-AB+B^{2})$

where $A = x^{2}, B = 1$ ;

$x^{6} +1 =\left ( x^{2} \right ) ^{3}+1^{3}$

$= (x^{2}+1)((x^{2})^{2}-x^{2} \cdot 1 + 1^{2})$

$= (x^{2}+1)(x^{4}-x^{2} + 1)$

The first factor,as the sum of squares, is a prime.

We try to factor the second by noting that it is "quadratic-style" based on $x^{2}$ . and can be written as

$= (x^{2})^{2}-x^{2} + 1$ ;

we seek to factor it as $(x^{2}+; ; ; )(x^{2}+; ; ; )$

We want a pair of integers whose product is 1 and whose sum is . These integers do not exist, so $x^{4}-x^{2} + 1$ is a prime.

$(x^{2}+1)(x^{4}-x^{2} + 1)$ is the prime factorization and the correct response is $x^{2}+1$ .

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Question

Which of the following is a prime factor of $n^{4} - 16n^{2} - 225$ ?

Answer

This can be most easily solved by first substituting for $n^{2}$ , and, subsequently, $t^{2}$ for $n^{4}$ :

$n^{4} - 16n^{2} - 225$

$= t^{2} - 16t - 225$

This becomes quadratic in the new variable, and can be factored as

$\left (t +; ; \right )\left (t +; ; \right )$ ,

filling out the blanks with two numbers whose sum is and whose product is . Through some trial and error, the numbers can be seen to be .

Therefore, after factoring and substituting back,

$n^{4} - 16n^{2} - 225$

$= t^{2} - 16t - 225$

$= (n^{2}+9)(n^{2}-25)$

The first factor, the sum of squares, is prime. The second factors as the difference of squares, so the final factorization is

$(n^{2}+9)(n+5)(n-5)$ .

Of the choices given, $n^{2}+9$ is correct.

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Question

Which of the following is a prime factor of $n^{4} + 30n^{2} +225$ ?

Answer

$n^{4} + 30n^{2} +225$ can be seen to fit the pattern

$A ^{2} + 2AB + B^{2}$ :

$n^{4} + 30n^{2} +225 = \left (n^{2} \right ) ^{2}+ 2 \cdot n^{2} \cdot 15 + 15 ^{2}$

where $A = n^{2} , B = 15$

$A ^{2}+2AB + B^{2}$ can be factored as $(A+B ) ^{2}$ , so

$n^{4} + 30n^{2} +225 = \left (n^{2} \right ) ^{2}+ 2 \cdot n^{2} \cdot 15 + 15 ^{2} = (n^{2}+15)^{2}$ .

$n^{2}+15$ does not fit into any factorization pattern, so it is prime, and the above is the complete factorization of the polynomial. Therefore, $n^{2}+15$ is the correct choice.

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Question

A triangle has a base of ft and height of ft. What is the area (in square feet) of the triangle?

Answer

The area of a triangle is: $A= \frac{1}{2}bh$

Use the FOIL Method to simplify.

$=\frac{1}{2} (x-2)(2x+2)$

$= \frac{1}{2}((x)(2x)) + (x)(2)+ (-2)(2x)-4)$

$= \frac{1}{2}(2x^2+2x-4x-4)$

$= \frac{1}{2}(2x^2-2x-4)$

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Question

Express 286 in base five.

Answer

To convert a base ten number to base five, divide the number by five, with the remainder being the digit in the units place; continue, dividing each successive quotient by five and putting the remainder in the next position to the left until the final quotient is less than five.

$286\div 5 = 57 \textrm{ R }\underline{1}$ - 1 is the last digit.

$57 \div 5 = 11 \textrm{ R }\underline{2}$ - 2 is the second-to-last digit.

$11 \div 5 = \underline{2} \textrm{ R }\underline{1}$ - 1 is the third-to-last digit; 2 is the first digit.

286 is equal to $2121_{\textrm{five}}$ .

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Question

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

Answer

A number increased by 20% is equivalent to 100% of the number plus 20% of the number. This is taking 120% of the number, or, equivalently, multiplying it by 1.2.

Therefore, increased by 20% is 1.2 times this, or

$1.2 \left (8x + 9y \right ) = 1.2 \cdot 8x + 1.2 \cdot 9y = 9.6x + 10.8 y$ .

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Question

Divide:

$\left ( 12 x ^{5} + 9x^{3}+ 6x^{2}+36 x \right ) \div 6x$

Answer

Divide termwise:

$\left ( 12 x ^{5} + 9x^{3}+ 6x^{2}+36 x \right ) \div 6x$

$= \frac{ 12 x ^{5} + 9x^{3}+ 6x^{2}+36 x }{6x}$

$= \frac{ 12 x ^{5} }{6x} + \frac{ 9x^{3} }{6x} + \frac{ 6x^{2} }{6x}+ \frac{ 36 x }{6x}$

$= \frac{ 12 }{6} x ^{5-1} + \frac{ 9 }{6 }x ^{3-1} + \frac{ 6 }{6}x^{2-1}+ \frac{ 36 }{6 }$

$=2x ^{4} + \frac{ 3 }{2 }x ^{2} + x+ 6$

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Question

Multiply:

$\left (100Y^{2}+ 110Y + 121 \right )(10Y-11)$

Answer

$\left (100Y^{2}+ 110Y + 121 \right )(10Y-11)$

$=\left [\left (10Y \right) ^{2}+ 10Y \cdot 11 + 11^{2} \right] (10Y-11)$

This product fits the difference of cubes pattern, where :

$(A^{2}+AB +B^{2})(A-B) = A^{3}-B^{3}$

so

$\left [\left (10Y \right) ^{2}+ 10Y \cdot 11 + 11^{2} \right] (10Y-11)$

$= \left ( 10Y\right )^{3} -11^{3} = 1,000Y^{3}-1,331$

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Question

Give the value of that makes the polynomial $121x^{2}- Nx + 49$ the square of a linear binomial.

Answer

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

$A^{2} \pm 2AB + B^{2}$ for some values of and .

$121x^{2}- Nx + 49= (11x)^{2} - Nx + 7^{2}$ , so

and .

For $121x^{2}- Nx + 49$ to be a perfect square, it must hold that

$Nx = 2AB = 2 \cdot 11x \cdot 7 = 154x$ ,

so . This is the correct choice.

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Question

Simplify the following:

Answer

Group all like terms by their order:

Simplify:

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Question

Simplify the following:

Answer

This can be solved using the FOIL method. The steps are shown below.

Therefore, after reordering, the answer is:

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