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Mathematics

1 answer:

goblinko [34]4 years ago

3 0

Rate of change is slope
So change in y/change in x
150/3=50

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What is the slope for y=7/2x - 2?

barxatty [35]

Easy peasy
slope intercept form is y=mx+b where m=slope

we see taht we have y=7/2x-2
therefor m=slope=7/2
the slope is 7/2

3 0

3 years ago

A firework rocket consists of a cone stacked on top of a cylinder, where the radii of the cone and the cylinder are equal. The d

Svet_ta [14]

Answer:

188π in^2

Step-by-step explanation:

Surface area of the rocket = surface area of the cone + surface area of the cylinder

Surface area of a cone = πr(r+h^2+r^2)

r = 8/2 = 4 in

4π( 4 + 3^2 + 4^2) =

4π ( 4 + 9 + 16) = 116π in^2

Surface area of a cylinder = 2πrh+2πr^2

(2 x n x 4 x 5) + (2 x n x 16)= 72π in^2

116 + 72 = 188π in^2

4 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:

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