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taurus [48]
3 years ago
14

Need help in b and c. show calculation pls.​

Mathematics
1 answer:
stellarik [79]3 years ago
8 0

√(16 - x^2) is defined only for -4 ≤ x ≤ 4, and is continuous over this domain, so

\displaystyle\lim_{x\to-4^+}\sqrt{16-x^2}=\sqrt{16-(-4)^2}=0

From the other side, the limit does not exist, because all x < -4 do not belong to the domain.

Taken together, the two-sided limit also does not exist.

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What is the value of the function y=2x−3 when x=−1 ?
san4es73 [151]

Answer:  First option is correct.

Step-by-step explanation:

Since we have given that

y=2x-3

We need to find the value of the function .

As we have given that x= - 1

so, we substitute the value of x in y :

y=2x-3\\\\y=2(-1)-3\\\\y=-2-3\\\\y=-5

So, the value of the function is -5.

Hence, First option is correct.


7 0
4 years ago
Read 2 more answers
Rationalize the denominator<br><br> 11/Square root of 5 - square root of 6
olasank [31]

Answer:

-11\sqrt{5}-11\sqrt{6}

Step-by-step explanation:

The given radical expression is;

\frac{11}{\sqrt{5}-\sqrt{6}}

We multiply both the numerator and the denominator by the conjugate of

\sqrt{5}-\sqrt{6} which is \sqrt{5}+\sqrt{6}

\frac{11(\sqrt{5}+\sqrt{6})}{(\sqrt{5}-\sqrt{6})(\sqrt{5}+\sqrt{6})}

The denominator is now difference of two squares;

\frac{11(\sqrt{5}+\sqrt{6})}{(\sqrt{5})^2-(\sqrt{6})^2}

Simplify:

\frac{11(\sqrt{5}+\sqrt{6})}{5-6}

\frac{11(\sqrt{5}+\sqrt{6})}{-1}

-11(\sqrt{5}+\sqrt{6})

-11\sqrt{5}-11\sqrt{6}

8 0
4 years ago
What was tonights homework for math?
Mila [183]
Well, looks like your out of luck cause we are not your teacher and most likely we are not your classmates, but hope you don't fail bud
4 0
3 years ago
What does it mean in this situation if the average rate of change is​ negative?
WARRIOR [948]
It means that it is a rate that is decreasing or declining.
5 0
4 years ago
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I need help please!!!!?
evablogger [386]

The answer would be n<4 because 4 is greater than n.

7 0
3 years ago
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