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3 years ago

12

Find the x and y intercepts of the line with equation 3y - 6 = 3?

2 answers:

storchak [24]3 years ago

4 0

The answer is 3 times y minus six equals to 3 .y=5 so three times five is fifteen so then you take that total and subtract it with six which equals to three . hope this helps you

Katyanochek1 [597]3 years ago

3 0

<u>X - Intercept</u>
3y - 6x = 3
3(0) - 6x = 3
    0 - 6x = 3
<u>-6x</u> = <u>3</u>
   -6 -6
   x = -0.5

<u>Y - Intercept</u>
3y - 6x = 3
3y - 6(0) = 3
    3y - 0 = 3
   <u>3y</u> = <u>3</u>
      3    3
   y = 1
<u>
X - Intercept and Y - Intercept</u>
(x, y) = (-0.5, 1)

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$m=\frac{9}{13}$ and $b=\frac{40}{13}$

Step-by-step explanation:

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We need to find the equation of the tangent line to the curve at the point (-3, 1).

Differentiate with respect to x.

$2[2(x^2+y^2)\frac{d}{dx}(x^2+y^2)]=25(2x-2y\frac{dy}{dx})$

$4(x^2+y^2)(2x+2y\frac{dy}{dx})=25(2x-2y\frac{dy}{dx})$

The point of tangency is (-3,1). It means the slope of tangent is $\frac{dy}{dx}_{(-3,1)}$ .

Substitute x=-3 and y=1 in the above equation.

$4((-3)^2+(1)^2)(2(-3)+2(1)\frac{dy}{dx})=25(2(-3)-2(1)\frac{dy}{dx})$

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$-240+80\frac{dy}{dx})=-150-50\frac{dy}{dx}$

$80\frac{dy}{dx}+50\frac{dy}{dx}=-150+240$

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Divide both sides by 130.

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If a line passes through a points $(x_1,y_1)$ with slope m, then the point slope form of the line is

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The slope of tangent line is $\frac{9}{13}$ and it passes through the point (-3,1). So, the equation of tangent is

$y-1=\frac{9}{13}(x-(-3))$

$y-1=\frac{9}{13}(x)+\frac{27}{13}$

Add 1 on both sides.

$y=\frac{9}{13}(x)+\frac{27}{13}+1$

$y=\frac{9}{13}(x)+\frac{40}{13}$

Therefore, $m=\frac{9}{13}$ and $b=\frac{40}{13}$ .

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3 years ago

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Answer:

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Step-by-step explanation:

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4 years ago

Solve for x: -3 + x = -22

Tanya [424]

Answer:

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Step-by-step explanation:

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3 years ago