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nadya68 [22]
3 years ago
7

Evaluate the limit, if it exists. Show work. lim→5 2−3−10 / 2−10

Mathematics
1 answer:
hram777 [196]3 years ago
4 0

Answer:

Convert the mixed numbers to improper fractions, then find the LCD and combine.

Exact Form: 13/3

Decimal Form:  4.3

Mixed Number Form:4  1/3

Step-by-step explanation:

Hope this helps ;)

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Step-by-step explanation:

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Hope this helps!!

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Read 2 more answers
What’s is the answer
kvasek [131]

When you are multiplying an exponent directly into a number/variable with an exponent, you multiply the exponents together.

For example:

(x^{2} )^{3} = x^6

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


When you are multiplying a variable with an exponent by another variable with an exponent, you add the exponents together.

For example:

(x^{2} )(x^{3})=x^{5}

(x^{1} )(x^{2})=x^{3}


(\frac{(x^{-3})(y^{2})}{(x^{4})(y^{6})} )^{3}=\frac{(x^{-9})(y^{6})}{(x^{12})(y^{18})}

You multiply 3 into each exponent in the numerator and the denominator

\frac{(x^{-9})(y^{6})}{(x^{12})(y^{18})}= \frac{y^{6}}{(x^{9})(x^{12})(y^{18})}

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

\frac{y^{6}}{(x^{21})(y^{18})} = \frac{1}{(x^{21})(y^{12})}


When you have something like this:

\frac{x^{2}}{x^5}

You subtract the exponents together, so:

\frac{x^2}{x^5} = x^{2-5} = x^{-3} = \frac{1}{x^3}


Your answer is the second option

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Answer:

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Step-by-step explanation:

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X = 5

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