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3 years ago

Which is an equation in point-slope form for the given point and slope? Point:(2,-6) Slope:-3/4

1 answer:

7 0

The equation for point-slope form is

$y-y_{1}=m(x-x_{1})$

m being the slope, y1 being the second point in the (2, -6) and the x1 being the first point in the (2, -6).

So the equation would look like this (for point-slope):

$y-(-6)=- \frac{3}{4}(x-2) \\ or \\ y+6=- \frac{3}{4}(x-2)$

If you're looking for slope-intercept form you can simplify the equation:

$y+6=- \frac{3}{4}(x-2) \\ \\ y+6=- \frac{3}{4}x+ \frac{6}{4} \\ \\ y+6-6=- \frac{3}{4}x+ \frac{6}{4}-6 \\ \\ y=- \frac{3}{4}x- \frac{18}{4}$

If you're looking for the standard form of the equation, you can take the slope-intercept equation and turn it into standard form:

$y=- \frac{3}{4}x- \frac{18}{4} \\ 4y=4(- \frac{3}{4}x- \frac{18}{4}) \\ 4y=-3x-18 \\ 4y+3x=-3x-18+3x \\ 4y+3x=-18 \\ 3x+4y=-18$

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4 years ago

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3 years ago

The amount Cami raised during last year’s charity walk, $45.50, is StartFraction 7 over 10 EndFraction of the amount she raised

barxatty [35]

Answer:

The equation is given by:

$n = \frac{7}{10} \times 45.50$

Step-by-step explanation:

During last years's charity walk, she raised $45.50.

During this years walk, she raised n, which is 7/10 of this amount. So, the equation is given by:

$n = \frac{7}{10} \times 45.50$

3 0

3 years ago

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$