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3 years ago

1/2h + 9 = 17 solve for h

Mathematics

1 answer:

tamaranim1 [39]3 years ago

6 0

Answer:

1/2h+9=17

We move all terms to the left:

1/2h+9-(17)=0

Domain of the equation: 2h!=0

h!=0/2

h!=0

h∈R

We add all the numbers together, and all the variables

1/2h-8=0

We multiply all the terms by the denominator

-8*2h+1=0

Wy multiply elements

-16h+1=0

We move all terms containing h to the left, all other terms to the right

-16h=-1

h=-1/-16

h=1/16

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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.

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

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

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

Therefore:

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

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5 0

2 years ago

Anyone feel like helping me??

Digiron [165]

The angles of a triangle add up to 180 degrees

65 + 3x - 10 + 2x = 180
5x + 55 = 180
5x = 180 - 55
5x = 125
x = 125/5
x = 25 <====

m < B = 3x - 10 = 3(25) - 10 = 75 - 10 = 65 <== B is 65 degrees
m < C = 2x......= 2(25) = 50 <== C is 50

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