Solve for y y/p-q = y/p+q

What is the value of the expression Зр divided by8 + 4 when p is 32?

Otrada [13]

Answer:

-4

Step-by-step explanation:

Divide each side by '-1'.

p = -4

Simplifying

p = -4

4 0

2 years ago

Can someone help plsss!

Roman55 [17]

Well the y-intercept is +2

And the slope of the blue line is 1/2

So y=1/2x+2

Slope is rise/run (up then ((right)))

And the intercept is found where the line intercects the y line.

7 0

2 years ago

Read 2 more answers

Someone please help me

pentagon [3]

Answer:

Option 3 or Option 4.

Step-by-step explanation:

I have this doubt because actually, a rhombus with a right angle (one) is not necessarily a square. Some rhombuses have one right angle and are still not squares. Now, if the four angles are right, it is a square, but this only says one.

On other hand, a parallelogram with congruent diagonals isn't a square necessarily either. Rectangles have congruent diagonals and are parallelograms.

In case you only can answer with one, I'll say is option 3, hope it helps!

8 0

2 years ago

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:

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