Numbers and Operations on Numbers - HiSET

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Question

Multiply, and express the product in scientific notation:

$7,200,000 \times 5,000,000$

Answer

Convert 7,200,000 to scientific notation as follows:

Move the (implied) decimal point until it is immediately after the first nonzero digit (the 7). This required moving the point six units to the left:

$7,200,000. \Rightarrow 7.200000$

It follows that 7,200,000 can be rewritten as $7.2 \times 10^{6}$ .

By similar reasoning, 5,000,000 can be rewritten as $5 \times 10^{6}$ .

Thus,

$7,200,000 \times 5,000,000$

$= (7.2\times 10 ^{6} )\times (5 \times 10^{6} )$

Rearrange and regroup the expressions so that the powers of ten are together:

$= (7.2\times 5) \times (10 ^{6} \times 10^{6} )$

Multiply the numbers in front. Also, multiply the powers of ten by adding exponents:

$= 36 \times 10 ^{6+6}$

$= 36 \times 10 ^{12}$

In order for the number to be in scientific notation, the number in front must be between 1 and 10. An adjustment must be made by moving the implied decimal point in 36 one unit left. It follows that

$36 = 3.6 \times 10^{1}$

Therefore,

$36 \times 10 ^{12} = (3.6 \times 10^{1}) \times 10 ^{12}$

$= 3.6 \times 10^{1+12}$

$= 3.6 \times 10^{13}$ ,

the correct response.

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Question

Simplify:

$\sqrt{120}$

Answer

To simplify a radical expression, first find the prime factorization of the radicand, which is 120 here.

$120 = 2 \times 2 \times 2 \times 3 \times 5$

Pair up like factors, then apply the Product of Radicals Property:

$\sqrt{120} =\sqrt{ 2 \times 2 \times 2 \times 3 \times 5} =\sqrt{ 2 \times 2 } \times \sqrt{ 2 \times 3 \times 5} = 2 \sqrt{30}$ ,

the simplest form of the radical.

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Question

Simplify:

$\sqrt{40}$

Answer

To simplify a radical expression, first find the prime factorization of the radicand, which is 40 here.

$40 = 2 \times 2 \times 2 \times 5$

Pair up like factors, then apply the Product of Radicals Property:

$\sqrt{40 }= \sqrt{2 \times 2 \times 2 \times 5} = \sqrt{2 \times 2 } \times \sqrt{ 2 \times 5} = 2 \sqrt{10}$ ,

the simplest form of the radical.

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Question

Simplify:

$\sqrt{32}$

Answer

To simplify a radical expression, first find the prime factorization of the radicand, which is 32 here.

$32 = 2 \times 2 \times 2 \times 2 \times 2$

Pair up like factors, then apply the Product of Radicals Property:

$\sqrt{32} = \sqrt{2 \times 2} \times\sqrt{ 2 \times 2} \times \sqrt{2 }= 2 \times 2 \times \sqrt{2 } = 4 \sqrt{2 }$ ,

the simplest form of the radical.

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Question

Simplify:

$\sqrt{66}$

Answer

To simplify a radical expression, first find the prime factorization of the radicand, which is 66 here.

$66 = 2 \times 3 \times 11$

$\sqrt{66} =\sqrt{ 2 \times 3 \times 11}$

There are no repeated prime factors, so the expression is already in simplified form.

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Question

Simplify the difference:

$\sqrt{70} - \sqrt{30}$

Answer

To simplify a radical expression, first, find the prime factorization of the radicand. First, we will attempt simplify $\sqrt{70}$ as follows:

$70 = 2 \times 5 \times 7$

Since no prime factor appears twice, the expression cannot be simplified further. The same holds for $\sqrt{30}$ , since $30 = 2 \times 3 \times 5$ .

It follows that the expression $\sqrt{70} - \sqrt{30}$ is already in simplest form.

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Question

Simplify the difference:

$\sqrt{176} - \sqrt{11}$

Answer

To simplify a radical expression, first find the prime factorization of the radicand. First, we will simplify $\sqrt{176}$ as follows:

$176 = 2 \times 2 \times 2 \times 2 \times 11$

By the Product of Radicals Property,

$\sqrt{176 }$

$= \sqrt{ 2 \times 2 \times 2 \times 2 \times 11}$

$= \sqrt{2 \times 2} \times \sqrt{2 \times 2} \times \sqrt{11 }$

$=2 \times 2 \times \sqrt{11 }$

$= 4 \sqrt{11}$

11 is a prime number, so $\sqrt{11}$ cannot be simplified. We can replace it with $1\sqrt{11}$ .

Through substitution, and the distribution property:

$\sqrt{176} - \sqrt{11}$

$= 4 \sqrt{11} -1 \sqrt{11}$

$=( 4 -1 )\sqrt{11}$

$=3\sqrt{11}$

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Question

Simplify the sum:

$\sqrt{99}+ \sqrt{44}+ \sqrt{11}$

Answer

To simplify a radical expression, first find the prime factorization of the radicand. First, we will simplify $\sqrt{99}$ as follows:

$99 = 3 \times 3 \times 11$

Pair up like factors, then apply the Product of Radicals Property.

$\sqrt{99} =\sqrt{ 3 \times 3 \times 11} = \sqrt{ 3 \times 3} \times \sqrt{ 11} = 3\sqrt{ 11}$

By similar reasoning,

$44 = 2 \times 2 \times 11$

$\sqrt{44} =\sqrt{ 2 \times 2 \times 11} = \sqrt{ 2 \times 2} \times \sqrt{ 11} = 2\sqrt{ 11}$

11 is a prime number, so $\sqrt{11}$ cannot be simplified. We can replace it with $1\sqrt{11}$ .

Through substitution, and the distribution property:

$\sqrt{99}+ \sqrt{44}+ \sqrt{11}$

$= 3\sqrt{ 11} + 2\sqrt{ 11} + 1 \sqrt{ 11}$

$= (3 + 2 + 1 )\sqrt{ 11}$

$= 6\sqrt{ 11}$

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Question

Simplify the sum:

$\sqrt{11} + \sqrt{22}+ \sqrt{33}$

Answer

To simplify a radical expression, first find the prime factorization of the radicand. First, we will attempt simplify $\sqrt{33}$ as follows:

$33 = 3 \times 11$

Since no prime factor appears twice, the expression cannot be simplified further. The same holds for $\sqrt{22}$ , since $22 = 2 \times 11$ ; the same also holds for $\sqrt{11}$ , since 11 is prime.

It follows that the expression $\sqrt{11} + \sqrt{22}+ \sqrt{33}$ is already in simplest form.

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Question

Simplify the expression:

$\frac{10}{\sqrt{15}}$

Answer

An expression with a radical expression in the denominator is not simplified, so to simplify, it is necessary to rationalize the denominator. This is accomplished by multiplying both numerator and denominator by the given square root, $\sqrt{15}$ , as follows:

$\frac{10}{\sqrt{15}} = \frac{10 \times \sqrt{15}}{\sqrt{15} \times \sqrt{15}} = \frac{10 \sqrt{15}}{15}$

The expression can be simplified further by dividing the numbers outside the radical by greatest common factor 5:

$\frac{10 \sqrt{15}}{15} = \frac{(10 \div 5) \sqrt{15}}{15 \div 5} = \frac{2 \sqrt{15}}{3}$

This is the correct response.

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Question

Simplify the expression

$\frac{6}{\sqrt{12}}$

Answer

An expression with a radical expression in the denominator is not simplified, so to simplify, it is necessary to rationalize the denominator. This is accomplished by multiplying both numerator and denominator by the given square root, $\sqrt{12}$ , as follows:

$\frac{6}{\sqrt{12}} = \frac{6 \times \sqrt{12}}{\sqrt{12} \times \sqrt{12}} = \frac{6 \times \sqrt{12}}{12}$

$\sqrt{12}$ can be simplified by taking the prime factorization of 12, and taking advantage of the Product of Radicals Property.

$12 = 2\times 2 \times 3$ , so

$\sqrt{12} = \sqrt{2 \times 2 \times 3 } = \sqrt{2 \times 2 } \times \sqrt{ 3 } = 2\sqrt{ 3 }$

Returning to the original expression and substituting:

$\frac{6 \times \sqrt{12}}{12} = \frac{6 \times 2 \sqrt{3}}{12} = \frac{12 \sqrt{3}}{12} =\sqrt{3}$ ,

the correct response.

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Question

Consider the expression $\frac{2\sqrt{6} }{ 3 \sqrt{5}}$ .

To simplify this expression, it is necessary to first multiply the numerator and the denominator by:

Answer

When simplifying an fraction with a denominator which is the product of an integer and a square root expression, it is necessary to first rationalize the denominator. This is accomplished by multiplying both halves of the fraction by the square root expression. The correct response is therefore $\sqrt{5}$ .

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Question

Consider the expression $\frac{1+ \sqrt{2}}{3- \sqrt{2}}$ .

To simplify this expression, it is necessary to first multiply the numerator and the denominator by:

Answer

When simplifying a fraction with a denominator which is the sum or difference of an integer and a square root, it is necessary to first rationalize the denominator. This is accomplished by multiplying both halves of the fraction by the conjugate of the denominator—the result of changing the plus symbol to a minus symbol (or vice versa); therefore, both halves of the given expression must be multiplied by the conjugate of $3 - \sqrt{2}$ , which is $3 + \sqrt{2}$ .

$3 + \sqrt{2}$ is therefore the correct choice.

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Question

Multiply: $\left (2 + \sqrt{2x} \right )\left (2 - \sqrt{2x} \right )$

Answer

The two expressions comprise the sum and the difference of the same two expressions, and can be multiplied using the difference of squares formula:

$\left (2 + \sqrt{2x} \right )\left (2 - \sqrt{2x} \right )$

$= 2^{2} - (\sqrt{2x} )^{2}$

The square of the square root of an expression is the expression itself, so:

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Question

Multiply: $(6 + \sqrt{21} )(6 - \sqrt{21} )$

Answer

The two expressions comprise the sum and the difference of the same two expressions, and can be multiplied using the difference of squares formula:

$(6 + \sqrt{21} )(6 - \sqrt{21} )$

$= 6^{2} -( \sqrt{21} )^{2}$

The square of the square root of an expression is the expression itself, so:

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Question

Multiply: $(7+2 \sqrt{7})(7-2 \sqrt{7})$

Answer

The two expressions comprise the sum and the difference of the same two expressions, and can be multiplied using the difference of squares formula:

$(7+2 \sqrt{7})(7-2 \sqrt{7})$

$= 7^{2} -(2 \sqrt{7})^{2}$

The second expression can be rewritten by the Power of a Product Property:

$= 7^{2} -2^{2}( \sqrt{7})^{2}$

The square of the square root of an expression is the expression itself:

By order of operations, multiply, then subtract:

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Question

Multiply: $\left ( \sqrt{7x}-\sqrt{3x} \right )\left ( \sqrt{7x}+\sqrt{3x} \right )$ .

Answer

The two expressions comprise the sum and the difference of the same two expressions, and can be multiplied using the difference of squares formula:

$\left ( \sqrt{7x}-\sqrt{3x} \right )\left ( \sqrt{7x}+\sqrt{3x} \right )$

$= ( \sqrt{7x})^{2}-(\sqrt{3x} )^{2 }$

The square of the square root of an expression is the expression itself:

By distribution:

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Question

Solve the following problem:

$\left ( \frac{(3+2-1\times4)}{4} \right )+\left ( (2+3)\times(4-2) \right )-((8\times 3)-2^3+(1+1))$

Answer

In order to solve this problem we need to follow the correct order of operations. Order of operations are outlined by the acronym PEMDAS: parenthesis, exponents, multiplication, division, addition, and subtraction. These operations should be performed in this order from the left to the right.

$\left ( \frac{(3+2-1\times4)}{4} \right )+\left ( (2+3)\times(4-2) \right )-((8\times 3)-2^3+(1+1))$

Let's start by solving the innermost parenthesis remember to work on exponents first.

$=\frac{(3+2-4)}{4} +\left ( 5\times2 \right )-(24-8+2)$

Simplify.

$=\frac{1}{4} +10-18$

$=10\tfrac{1}{4}-18$

$=-7\tfrac{3}{4}$

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Question

Rewrite the expression $\sqrt{x} \cdot \sqrt[3]{x}$ as a single radical expression.

Answer

First, rewrite each radical expression using rational exponents:

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

$\sqrt[3]{x} = x ^{\frac{1}{3}}$

Multiply these expressions using the Product of Powers Property:

$\sqrt{x} \cdot \sqrt[3]{x}$

$= x ^{\frac{1}{2}} \cdot x ^{\frac{1}{3}}$

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

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

$= x ^{\frac{5}{6}}$

This can be rewritten in radical form as $\sqrt[6]{x^{5}}$ .

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Question

Rewrite as a single power of :

$(y^{2})^{3} \cdot y (y^{2})$

Answer

Apply the properties of exponents to the expression as follows:

$(y^{2})^{3} \cdot y (y^{2})$

By the Power of a Power Property:

$= (y^{2 \cdot 3})\cdot y (y^{2})$

$= (y^{6})\cdot y (y^{2})$

Restating:

$= y^{6} \cdot y^{1} \cdot y^{2}$

By the Product of Powers of Property:

$= y^{6+1+2}$

$= y^{9}$ ,

the correct response.

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