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2 years ago

Write the point-slope equation oft line that goes through (1, -1) with slope of -5.

Mathematics

1 answer:

Sever21 [200]2 years ago

6 0

Answer:

y+1=-5(x-1)

Step-by-step explanation:

We need to find the slope intercept form first. So far we have this: y=-5x+b. We could plug in the x value and y value to get this : -1=-5(1)+b. b=4. Now we have the slope intecept form: y=-5x+4, and we could turn it into point slope form.

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Solve the simultaneous equation<br> X+3y=13<br> X-y=5

Andru [333]

Answer:

x = 7

y = 2

Step-by-step explanation:

In the above question, we are given 2 equations which are simultaneous. To solve this equation, we have to find the values of x and y

x + 3y = 13 ........ Equation 1

x - y = 5...........Equation 2

From Equation 2,

x = 5 + y

Substitute 5 + y for x in Equation 1

x + 3y = 13 ........ Equation 1

5 + y + 3y = 13

5 + 4y = 13

4y = 13 - 5

4y = 8

y = 8/4

y = 2

Since y = 2, substitute , 2 for y in Equation 2

x - y = 5...........Equation 2

x - 2 = 5

x = 5 + 2

x = 7

Therefore, x = 7 and y = 2

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3 years ago

Please help with this problem!

Dovator [93]

Answer:

x+14

Step-by-step explanation:

(12+x)+2

12+x+2

x+(12+2)

x+14

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3 years ago

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In which quadrant does 694 lie?

egoroff_w [7]

Answer: C. quadrant l

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3 years ago

What is the y-intercept of

liberstina [14]

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2 years ago

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I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur

Butoxors [25]

$\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx$

Setting $x=5\sin\theta$ , you have $\mathrm dx=5\cos\theta\,\mathrm d\theta$ . Then the integral becomes

$\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta$
$\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

Now, $\sqrt{x^2}=|x|$ in general. But since we want our substitution $x=5\sin\theta$ to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means $\theta=\sin^{-1}\dfrac x5$ , which implies that $\left|\dfrac x5\right|\le1$ , or equivalently that $|\theta|\le\dfrac\pi2$ . Over this domain, $\cos\theta\ge0$ , so $\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta$ .

Long story short, this allows us to go from

$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

to

$\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta$

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

$\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta$

Then integrate term-by-term to get

$\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)$
$=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C$

Now undo the substitution to get the antiderivative back in terms of $x$ .

$=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C$

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

$=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C$

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2 years ago

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Write the point-slope equation oft line that goes through (1, -1) with slope of -5.​

Write the point-slope equation oft line that goes through (1, -1) with slope of -5.