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GarryVolchara [31]
2 years ago
6

Find the equation of the line shown.

Mathematics
2 answers:
vladimir1956 [14]2 years ago
4 0

Answer:

y = - x + 9

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

calculate m using the slope formula

m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁ ) = (0, 9) and (x₂, y₂ ) = (9, 0) ← 2 points on the line

m = \frac{0-9}{9-0} = \frac{-9}{9} = - 1

the line crosses the y- axis at (0, 9 ) ⇒ c = 9

y = - x + 9 ← equation of line

I am Lyosha [343]2 years ago
3 0

Answer:

y = -x + 9

Step-by-step explanation:

<u>Slope</u>

  • m = rise / run
  • m = -1 / 1
  • m = -1

<u>Y-intercept</u>

  • (0, 9) = (0, b)
  • b = 9

<u>Equation</u>

  • y = mx + b
  • y = -x + 9
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Calculus, question 5 to 5a​
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\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - x^2}

and dx = \cos(\theta) \, d\theta. So the integral transforms to

\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = \int \frac{\sin^3(\theta)}{\cos(\theta)} \cos(\theta) \, d\theta = \int \sin^3(\theta) \, d\theta

Reduce the power by writing

\sin^3(\theta) = \sin(\theta) \sin^2(\theta) = \sin(\theta) (1 - \cos^2(\theta))

Now let y = \cos(\theta), so that dy = -\sin(\theta) \, d\theta. Then

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Replace the variable to get the antiderivative back in terms of x and we have

\displaystyle \int \frac{x^3}{\sqrt{1-x^2}} \, dx = -\cos(\theta) + \frac13 \cos^3(\theta) + C

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since, like in the previous integral, under this reversible variable change we assume -π/2 < θ < π/2. Over this interval, sec(θ) is positive.

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