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4 years ago

13

What function has a zero of -5

1 answer:

zepelin [54]4 years ago

8 0

There are infinitely many.

x+5=0
x^2-25=0
x^2+2x-15=0

again there are infinitely many

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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

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

6 0

3 years ago

Please help no links or random stuff!!

Dmitrij [34]

Answer:

See explanation. I added a photo and tried to explain everything. Please tell me in the comments if I explained it well and if handwriting was ok :)

Step-by-step explanation:

3 0

3 years ago

E + 1 13/16 = 2 5/16

Burka [1]

e + 1 13/16 = 2 5/16

subtract 1 13/16 from each side

e = 2 5/16 - 1 13/16

borrow from the 2

e = 1 16/16 + 5 /16 - 1 13/16

e = 1 21/16-1 13/16

e = 1 8 /16

e = 1 1/2

7 0

3 years ago

The radius of a circle is 4 inches what is the circles diameter

Leokris [45]

Answer:

The diameter of the circle is 8 inches

Step-by-step explanation:

The diameter is 2 times the radius

5 0

3 years ago

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natka813 [3]

Answer:

8

Step-by-step explanation:

6 0

3 years ago

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