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Anvisha [2.4K]
3 years ago
11

What is the exact area of a circle having diameter 7 in.?

Mathematics
1 answer:
grandymaker [24]3 years ago
6 0
Area of circle = <span>πr^2
r = 7/2 = 3.5
r^2 = 12.25

Area = 12.25</span><span>π in^2</span>
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Please help with only the circled ones (1-8)
mixas84 [53]

When you have an exponent divided by another exponent, you subtract the exponents (only when it has the same base)

For example:

\frac{x^8}{x^3} =x^{8-3}=x^5

\frac{x^3}{x^2} =x^{3-2}=x^{1}  


When you have a negative exponent, you move it to the other side of the fraction to make the exponent positive

For example:

x^{-2}=\frac{1}{x^2}

\frac{1}{x^{-5}}=\frac{x^5}{1} = x^5

\frac{y^{-2}}{x^{-1}} =\frac{x^1}{y^2} =\frac{x}{y^2}


1. \frac{10^{15}}{10^3} =10^{15-3} = 10^{12}


2. \frac{(-3)^4}{(-3)^{-3}} =(-3)^{4-(-3)}=(-3)^{4+3} = (-3)^7


3. \frac{8}{8^3} =8^{1-3} = 8^{-2}=\frac{1}{8^2}


4. \frac{a^{12}}{a^2} =a^{12-2}=a^{10}


5. \frac{m^{-2}n^{16}}{m^{4}n^2} =(m^{-2-4})(n^{16-2})=(m^{-6})(n^{14})=\frac{n^{14}}{m^{6}}

This is one of the ways you could have done it


6. \frac{p^5q^{-10}}{p^6q^{-2}} =(p^{5-6})(q^{-10-(-2)})=(p^{-1})(q^{-8})=\frac{1}{p^1q^8} =\frac{1}{pq^8}


7. \frac{63x^{18}}{9x^{2} }   Divide 63 and 9

\frac{7x^{18}}{x^{2}} =(7)(x^{18-2})=(7)(x^{16})=7x^{16}


8. \frac{28r^4}{-7r^{15}} =(\frac{28}{-7} )(r^{4-15})=(-4)(r^{-11})=(-4)(\frac{1}{r^{11}} )=\frac{-4}{r^{11}}


[More information with exponents]

If you multiply an exponent directly with another exponent, you multiply the exponents together

For example:

(x^{2})^4=x^{2(4)}=x^8

(x^{3})^5 =x^{3(5)}=x^{15}


If you multiply a variable with an exponent by a variable with an exponent, you add the exponents

For example:

(x^{2}) (x^6)=x^{2+6}=x^8

(x^{3})(x^1)=x^{3+1}=x^4

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