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2 years ago

U

Mathematics

2 answers:

True [87]2 years ago

8 0

Answer:

D. RWS and SWT

Step-by-step explanation:

Adjacent angles have a common side and a common vertex but they don't overlaps each other

The two angles that has these requirements are :

RWS and SWT

KIM [24]2 years ago

7 0

The two angles are said to be adjacent angles when they share the common vertex and side.

So , adjacent angles are :

Option D is the correct answer.

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Value of 3x^2 +4y^2 if x=2, y=1, and z= -3

Dennis_Churaev [7]

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3 years ago

Erify that f has an inverse. then use the function f and the given real number a to find (f −1)'(a). (hint: see example 1. if an

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The function

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3 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

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Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$