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mixer [17]
3 years ago
10

How do I go about solving this? Do I make it into one big fraction or can I simplify the radical somehow?

Mathematics
1 answer:
asambeis [7]3 years ago
5 0
Not sure how to find this limit algebraically.

However, you can evaluate it analytically.
Notice the domain is restricted to:
x^2 \ \textgreater \  4
This means the right sided limit cannot exist.
You can only evaluate the left side limit.

As x-> -2 from the left. The first term goes to positive infinity and 2nd term goes to negative infinity.

\infty - (-\infty) = \infty

Therefore the limit is positive infinity.
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Alexandra [31]

Answer:

Its H or J

I Would go with J tho...

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3 0
3 years ago
You download 12 news songs to your phone. Then you delete 5 old songs.Write each amount as an integer.
Gennadij [26K]
Hey there! :D

When you have a value, it is usually positive.

For example, if I have 5 chocolate bars, then I have +5 chocolate bars. 

When you take away a value, it is usually negative.

For example, I have 3 chocolate bars away, then I would have 3 fewer chocolate bars. (-3) 

Integers are negative or positive whole numbers.

12 songs= 12 <== the integer

Delete 5 old songs= -5 <=== the integer

I hope this helps!
~kaikers
8 0
3 years ago
Read 2 more answers
Janice listed the integers below on the chalkboard. Which of these integers has the greatest opposite? 1, –6, 11, –10, –8, 9, 5
muminat
-10 because its a negative and that's the opposite of the greatest itreger
8 0
4 years ago
Can someone plz help
erica [24]

Answer:

4th choice which is 50.24 sq.m

4 0
3 years ago
Read 2 more answers
5^(-x)+7=2x+4 This was on plato
Setler79 [48]

Answer:

Below

I hope its not too complicated

x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}

Step-by-step explanation:

5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}

Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}

\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x

\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}

3 0
3 years ago
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