−6∣x−2∣−9=4∣x−2∣+5 what are the solutions of the following equations?

The measure of a central angle is 3x + 18 and the measure of its arc is 147 which equation can you use to find the value of x

steposvetlana [31]

3x + 18 = 147

3x = 129

x = 43

7 0

2 years ago

What number must be added to the expression below to complete the square ? x2+7x

stellarik [79]

Answer:

you must add 2 to the equation

Step-by-step explanation:

6 0

3 years ago

Can I please get some help with both problems

dimulka [17.4K]

So you subtract 180 from 90 you get you answer then you will divide it by 55 and your remaining answer which you will get will be x

4 0

3 years ago

a homework assignment included 40 multiple choice questions Sabrina completed 65% of her assignment in David completed 4/5 of hi

bazaltina [42]

Answer:

6 or -6

Step-by-step explanation:

65% of 40=26
4/5 of 40=32
32-26=6
(or if you start with 26)
26-32=-6

7 0

2 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:

Fundamental Theorem of Calculus

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

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

$\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 indefinite integral:

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

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

1 year ago