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eimsori [14]
3 years ago
15

Plz help i need to get this done

Mathematics
1 answer:
netineya [11]3 years ago
8 0

Answer:

$14.40

-----------------

please give brainliest.

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the X represents the width, since to get the area, you have to multiply length times width

Hope that helps!

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It's a 7th grade math problem ​
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1296 cm²

Step-by-step explanation:

36² = 1296 cm²

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2 years ago
Aaron has two apple's. He adds two more apples. How much apple's does Aaron has now
lorasvet [3.4K]
Aaron has 4 apples. 2+2=4
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5/10 + 8/100<br> answer the problem
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How to solve this derivative.<br><br> f(x)={ln(3x)}^2x<br> with steps...
AysviL [449]

Rewrite <em>f(x)</em> as a nested exponential-logarithm expression :

\left(\ln(3x)\right)^{2x} = \exp\left(\ln\left(\ln(3x)\right)^{2x}\right)

(where \exp(x) = e^x)

One of the properties of logarithms lets us drop the exponent as a coefficient:

\exp\left(\ln\left(\ln(3x)\right)^{2x}\right) = \exp\left(2x\ln\left(\ln(3x)\right)\right)

Now, by the chain rule, we have

f(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \implies \\\\ f'(x) = \left(\exp\left(2x\ln\left(\ln(3x)\right)\right)\right)' \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2x\ln\left(\ln(3x)\right)\right)'

By the product rule,

f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(   (2x)' \ln\left(\ln(3x)\right) + 2x\left(\ln\left(\ln(3x)\right)\right)'   \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(   2\ln\left(\ln(3x)\right) + 2x\left(\ln\left(\ln(3x)\right)\right)'   \right)

By the chain rule again,

f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + 2x \cdot \dfrac1{\ln(3x)} \cdot \left(\ln(3x)\right)'  \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + \dfrac{2x}{\ln(3x)} \cdot \dfrac1{3x} \cdot (3x)' \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + \dfrac{2}{3\ln(3x)} \cdot 3 \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + \dfrac{2}{\ln(3x)} \right)

Then simplify this to

f'(x) = \boxed{\left(\ln(3x)\right)^{2x} \left(2\ln\left(\ln(3x)\right) + \dfrac{2}{\ln(3x)} \right)}

8 0
3 years ago
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