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Alex_Xolod [135]
3 years ago
10

Find the indicated limit, if it exists.(7 points)

Mathematics
1 answer:
Alecsey [184]3 years ago
7 0

Answer:

7

Step-by-step explanation:

The left hand limit is when we approach zero from left. We use the function on this domain in finding the limit.

\lim_{x \to 0^-} f(x)=7-x^2

\lim_{x \to 0^-} f(x)=7-(0)^2=7

The right hand limit is

\lim_{x \to 0^+} f(x)=10x+7

\lim_{x \to 0^+} f(x)=10(0)+7=7

Since the left hand limit equals the right hand limit;

\lim_{x \to 0} f(x)=7

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-3(x-4)+x=2x-12 what does x =
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Step-by-step explanation:

1)  Expand the equation to -2x+12=2x-12

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MatroZZZ [7]

For this case we must simplify the following expression:

\frac {6-3 \sqrt [3] {6}} {\sqrt [3] {9}}

Multiplying the numerator and denominator by(\sqrt [3] {9}) ^ 2

\frac {6-3 \sqrt [3] {6}} {\sqrt [3] {9}} * \frac {(\sqrt [3] {9}) ^ 2} {(\sqrt [3] { 9}) ^ 2} =

We rewrite:

\frac {\frac {6-3 \sqrt [3] {6}} * (\sqrt [3] {9}) ^ 2} {\sqrt [3] {9} * (\sqrt [3] {9 }) ^ 2} =

By properties of powers we have that:

a ^ m * a ^ n = a ^ {m + n}\\\frac {(6-3 \sqrt [3] {6}) * (\sqrt [3] {9}) ^ 2} {(\sqrt [3] {9}) ^ 3} =\\\frac {(6-3 \sqrt [3] {6}) * (\sqrt [3] {9}) ^ 2} {9} =

We rewrite, moving the exponent within the radical:

\frac {(6-3 \sqrt [3] {6}) * \sqrt [3] {9 ^ 2}} {9} =\\\frac {(6-3 \sqrt [3] {6}) * \sqrt [3] {81}} {9} =

We can rewrite3 * 3 ^ 3 = 81

\frac {(6-3 \sqrt [3] {6}) * \sqrt [3] {3 * 3 ^ 3}} {9} =

We simplify:

\frac {(6-3 \sqrt [3] {6}) * 3 \sqrt [3] {3}} {9} =

We apply distributive property:

\frac {18 \sqrt [3] {3} -9 \sqrt [3] {18}} {9} =

Simplifying we finally have:

2 \sqrt [3] {3} - \sqrt [3] {18}

Answer:

2 \sqrt [3] {3} - \sqrt [3] {18}

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