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fenix001 [56]
3 years ago
6

Plz help and ty very much​

Mathematics
1 answer:
IgorC [24]3 years ago
4 0

Answer:

True

Step-by-step explanation:

For a compound inequality using the word "or" to be true, all you need is for one of the parts to be true.

-6 < 7 is true

-2 <= -2 is also true

Answer: True

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Match the numerical expressions to their simplified forms
eduard

Answer:

1.\ \ p^2q = (\frac{p^5}{p^{-3}q^{-4}})^{\frac{1}{4}}

2.\ \ pq^{\frac{3}{2}}} = (\frac{p^2q^7}{q^{4}})^{\frac{1}{2}}

3.\ \ pq^2 = \frac{(pq^3)^{\frac{1}{2}}}{(pq)^{\frac{-1}{2}}}

4.\ \ p^2q^{\frac{1}{2}} = (p^6q^{\frac{3}{2}})^{\frac{1}{3}}

Step-by-step explanation:

Required

Match each expression to their simplified form

1.

(\frac{p^5}{p^{-3}q^{-4}})^{\frac{1}{4}}

Simplify the expression in bracket by using the following law of indices;

\frac{a^m}{a^n} = a^{m-n}

The expression becomes

(\frac{p^{5-(-3)}}{q^{-4}})^{\frac{1}{4}}

(\frac{p^{5+3}}{q^{-4}})^{\frac{1}{4}}

(\frac{p^8}{q^{-4}})^{\frac{1}{4}}

Split the fraction in the bracket

(p^8*\frac{1}{q^{-4}})^{\frac{1}{4}}

Simplify the fraction by using the following law of indices;

\frac{1}{a^{-m}} = a^m

The expression becomes

(p^8*q^4)^{\frac{1}{4}}

Further simplify the expression in bracket by using the following law of indices;

(ab)^m = a^m * b^m

The expression becomes

(p^{8*\frac{1}{4}}\ *\ q^4*^{\frac{1}{4}})

(p^{\frac{8}{4}}\ *\ q^{\frac{4}{4}})

p^2q

Hence,

(\frac{p^5}{p^{-3}q^{-4}})^{\frac{1}{4}} = p^2q

2.

(\frac{p^2q^7}{q^{4}})^{\frac{1}{2}}

Simplify the expression in bracket by using the following law of indices;

\frac{a^m}{a^n} = a^{m-n}

The expression becomes

({p^2q^{7-4}}})^{\frac{1}{2}}

({p^2q^3}})^{\frac{1}{2}}

Further simplify the expression in bracket by using the following law of indices;

(ab)^m = a^m * b^m

The expression becomes

{p^{2*\frac{1}{2}}q^{3*\frac{1}{2}}}}

pq^{\frac{3}{2}}}

Hence,

pq^{\frac{3}{2}}} = (\frac{p^2q^7}{q^{4}})^{\frac{1}{2}}

3.

\frac{(pq^3)^{\frac{1}{2}}}{(pq)^{\frac{-1}{2}}}

Simplify the numerator as thus:

\frac{p^{\frac{1}{2}} * q^3*^{\frac{1}{2}}}{(pq)^{\frac{-1}{2}}}

\frac{p^{\frac{1}{2}} * q^{\frac{3}{2}}}{(pq)^{\frac{-1}{2}}}

Simplify the denominator as thus:

\frac{p^{\frac{1}{2}} * q^{\frac{3}{2}}}{p^{\frac{-1}{2}}q^{\frac{-1}{2}}}

Simplify the expression in bracket by using the following law of indices;

\frac{a^m}{a^n} = a^{m-n}

The expression becomes

p^{\frac{1}{2} - (\frac{-1}{2} )} * q^{\frac{3}{2} - (\frac{-1}{2}) }

p^{\frac{1}{2} +\frac{1}{2} } * q^{\frac{3}{2} + \frac{1}{2} }

p^{\frac{1+1}{2}} * q^{\frac{3+1}{2}}

p^{\frac{2}{2}} * q^{\frac{4}{2}}

pq^2

Hence,

pq^2 = \frac{(pq^3)^{\frac{1}{2}}}{(pq)^{\frac{-1}{2}}}

4.

(p^6q^{\frac{3}{2}})^{\frac{1}{3}}

Simplify the expression in bracket by using the following law of indices;

(ab)^m = a^m * b^m

The expression becomes

p^6*^{\frac{1}{3}}\ *\ q^{\frac{3}{2}}*^{\frac{1}{3}}

p^{\frac{6}{3}}\ *\ q^{\frac{3*1}{2*3}}

p^2 *\ q^{\frac{3}{6}}

p^2 *\ q^{\frac{1}{2}

p^2q^{\frac{1}{2}

Hence

p^2q^{\frac{1}{2}} = (p^6q^{\frac{3}{2}})^{\frac{1}{3}}

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3 years ago
A stray dog ate 40 of your muffins. That was 5/6 of all them! How many are left?
lilavasa [31]
I think that there are 33 left.

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inessss [21]

Answer:

26

Step-by-step explanation:

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4 years ago
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The sum of two integers with different signs is eight give to possible integers that fit this description
Soloha48 [4]
An easy way is to do this
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8 0
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