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2 years ago

11

v A square garden has an area of 110 square feet. a. What is the approximate length of one side of the garden? Round to the near

est tenth.

1 answer:

tankabanditka [31]2 years ago

8 0

Idkdkdndmnebebebwbwwjwo

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Find the sale price to the nearest cent<br> $53 pair of jeans with a 20% discount

Artist 52 [7]

Answer: It is $42.4

Step-by-step explanation: Multiply (0.20)(53), you get 10.6, now that's the price that is discounted, so subtract that from the total, to get $42.4. If you want to check your answer, do 42.4/53 and you would get 0.80, or 80%. That is what you should get since you have a 20% discount from 100% of a product.

8 0

2 years ago

Is x 12 or 15? Or none?

Usimov [2.4K]

The answer is 15 buddy!

7 0

3 years ago

Read 2 more answers

What is 5047 divided by 52?

xenn [34]

Answer:

97.06

Step-by-step explanation:

5047 / 52

= 97.06

4 0

3 years ago

Read 2 more answers

Find integra of xlnxdx

Mademuasel [1]

Answer:

$\dfrac{1}{2}x^2\ln x - \dfrac{1}{4}x^2+\text{C}$

Step-by-step explanation:

<u>Fundamental Theorem of Calculus</u>

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a <u>constant of integration</u>.

$\boxed{\begin{minipage}{4 cm}\underline{Integrating $x^n$}\\\\$\displaystyle \int x^n\:\text{d}x=\dfrac{x^{n+1}}{n+1}+\text{C}$\end{minipage}}$

Increase the power by 1, then divide by the new power.

Given <u>indefinite integral</u>:

$\displaystyle \int x \ln x \:\: \text{d}x$

To integrate the given integral, use Integration by Parts:

$\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}$

$\text{Let }u=\ln x \implies \dfrac{\text{d}u}{\text{d}x}=\dfrac{1}{x}$

$\text{Let }\dfrac{\text{d}v}{\text{d}x}=x \implies v=\dfrac{1}{2}x^2$

Therefore:

$\begin{aligned}\displaystyle \int u \dfrac{dv}{dx}\:dx & =uv-\int v\: \dfrac{du}{dx}\:dx\\\\\implies \displaystyle \int x \ln x\:\:\text{d}x & = \ln x \cdot \dfrac{1}{2}x^2-\int \dfrac{1}{2}x^2 \cdot \dfrac{1}{x}\:\:dx\\\\& = \dfrac{1}{2}x^2\ln x -\int \dfrac{1}{2}x\:\:dx\\\\& = \dfrac{1}{2}x^2\ln x - \dfrac{1}{4}x^2+\text{C}\end{aligned}$

Learn more about integration here:

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5 0

2 years ago

Your current schedule only lets you take two 1-credit courses every six months. You need 12 credits to get the degree you want.

lianna [129]

So,

Since you can only take two one-credit courses every six months, you can earn 2 credits every six months, and 4 credits in one year. Multiplying that by 3, you get 12 credits in 3 years.

It will take 3 years to get the degree.

6 0

3 years ago