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3 years ago

What is square root of 29

Mathematics

1 answer:

VladimirAG [237]3 years ago

8 0

Answer:

5.38516480713

Step-by-step explanation:

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Solve the following for y:<br> p(t + 4) = 2y

dimulka [17.4K]

Answer:

the answer is y =p(t + 4)/2

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3 years ago

Solve r = (9–9) for p.

Dmitry_Shevchenko [17]

Answer: u mean r=0

Step-by-step explanation:

5 0

3 years ago

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Differentiate x^2 e^x logx

zimovet [89]

Product rule:

$\dfrac{\mathrm d}{\mathrm dx}(x^2e^x\log x)$

$=\dfrac{\mathrm d(x^2)}{\mathrm dx}e^x\log x+x^2\dfrac{\mathrm d(e^x)}{\mathrm dx}\log x+x^2e^x\dfrac{\mathrm d(\log x)}{\mathrm dx}$

Power rule:

$\dfrac{\mathrm d(x^2)}{\mathrm dx}=2x$

The exponential function is its own derivative:

$\dfrac{\mathrm d(e^x)}{\mathrm dx}=e^x$

Assuming the base of $\log x$ is $e$ , its derivative is

$\dfrac{\mathrm d(\log x)}{\mathrm dx}=\dfrac1x$

But if you mean a logarithm of arbitrary base $b$ , we have

$y=\log_bx\implies x=b^y=e^{y\ln b}\implies1=e^{y\ln b}\ln b\dfrac{\mathrm dy}{\mathrm dx}$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{e^{-y\ln b}}{\ln b}=\dfrac1{b^y\ln b}$

$\implies\dfrac{\mathrm d(\log_bx)}{\mathrm dx}=\dfrac1{x\ln b}$

So we end up with

$2xe^x\log x+x^2e^x\log x+\dfrac{x^2e^x}x$

$=xe^x(2\log x+x\log x+1)$

8 0

3 years ago

What is w-4>9 written on a number line?

Aleksandr-060686 [28]

There would be an open circle on positive 12 with the arrow moving to the right.

7 0

3 years ago

Are these correct? If not please help me write the correct answer.

Keith_Richards [23]

Answer:

Mostly correct

For Q3, you are doing great, you remembered that negative numbers can happen if the exponent is an odd number.

For Q4, I think you were confused with the fractions,

4(i)

$(\frac{2}{3} )^{4} = (\frac{2^{4}}{3^{4}}) = (\frac{16}{81}) =0.198$

4(ii) is also wrong but I'll let you try to fix it yourself following what I given you for 4(i). You should get 1/64

Hope this helps!

4 0

3 years ago

Read 2 more answers