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2 years ago

A line intersects the points (-1, 3) and

Mathematics

1 answer:

olya-2409 [2.1K]2 years ago

4 0

Answer:

I was able to get: y = -(6/3)x + 1

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Find the value of H:

rewona [7]

Answer:

The value of H is 28.

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(H-6)/(-7-(-5))

m=(H-6)/(-7+5)

-11=(H-6)/-2

-11(-2)=H-6

22=H-6

H=22+6

H=28

7 0

2 years ago

- {8, 2.1, 145, 178, 3} which one is irrational

ratelena [41]

Answer:

These numbers are not irrational .

Step-by-step explanation:

They are all whole numbers with no decimal points and ongoing numbers.

4 0

2 years ago

At a certain time of day, the angle of elevation of the sun is 40°. To the nearest foot,find the height of a tree whose shadow i

MrRa [10]

Answer: approximately 29 feet

Explanation: You need to find a tree so that the angle of elevation from the end of the shadow to top of the tree is 40 degrees.

The length of the shadow is an adjacent side and is 35.
The height of the tree is the opposite side. You could use X.

Tan ratio = opposite/adjacent

tan(40) = x/35
x = 35*tan(40) =29.37

5 0

3 years ago

We want to solve the following system of equations using the substitution method.

inna [77]

In the second equation you are given what Y equals, which is (-5x - 3). You would use this equation and plug it into the y value given in the first equation where it says 2y and solve

That would be

3x - 2(-5x - 3) = -6
3x + 10x + 6 = -6
3x + 10x = -6 - 6
13x = -12
X = -12/13

Then if you want to solve for Y you can use any equation and plug in the x-value found.

I’m going to use equation 2.

Y = -5x - 3
Y = -5(-12/13) - 3
Y = 4.615 - 3
Y = 1.615

(-12/13, 1.615)

Therefore the x-value is -12/13 and the y-value is 1.615.

5 0

2 years ago

Integrate sin^-1(x) dx please explain how to do it aswell ...?

Lynna [10]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2264253

_______________

Evaluate the indefinite integral:

$\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx\qquad\quad\checkmark}$

Trigonometric substitution:

$\mathsf{\theta=sin^{-1}(x)\qquad\qquad\dfrac{\pi}{2}\le \theta\le \dfrac{\pi}{2}}$

then,

$\begin{array}{lcl} \mathsf{x=sin\,\theta}&\quad\Rightarrow\quad&\mathsf{dx=cos\,\theta\,d\theta\qquad\checkmark}\\\\\\ &&\mathsf{x^2=sin^2\,\theta}\\\\ &&\mathsf{x^2=1-cos^2\,\theta}\\\\ &&\mathsf{cos^2\,\theta=1-x^2}\\\\ &&\mathsf{cos\,\theta=\sqrt{1-x^2}\qquad\checkmark}\\\\\\ &&\textsf{because }\mathsf{cos\,\theta}\textsf{ is positive for }\mathsf{\theta\in \left[\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right].} \end{array}$

So the integral $\mathsf{(ii)}$ becomes

$\mathsf{=\displaystyle\int\! \theta\,cos\,\theta\,d\theta\qquad\quad(ii)}$

Integrate $\mathsf{(ii)}$ by parts:

$\begin{array}{lcl} \mathsf{u=\theta}&\quad\Rightarrow\quad&\mathsf{du=d\theta}\\\\ \mathsf{dv=cos\,\theta\,d\theta}&\quad\Leftarrow\quad&\mathsf{v=sin\,\theta} \end{array}\\\\\\\\ \mathsf{\displaystyle\int\!u\,dv=u\cdot v-\int\!v\,du}\\\\\\ \mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta-\int\!sin\,\theta\,d\theta}\\\\\\ \mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta-(-cos\,\theta)+C}$

$\mathsf{\displaystyle\int\!\theta\,cos\,\theta\,d\theta=\theta\, sin\,\theta+cos\,\theta+C}$

Substitute back for the variable x, and you get

$\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx=sin^{-1}(x)\cdot x+\sqrt{1-x^2}+C}\\\\\\\\ \therefore~~\mathsf{\displaystyle\int\!sin^{-1}(x)\,dx=x\cdot\,sin^{-1}(x)+\sqrt{1-x^2}+C\qquad\quad\checkmark}$

I hope this helps. =)

Tags: integral inverse sine function angle arcsin sine sin trigonometric trig substitution differential integral calculus

6 0

2 years ago