Use the given conditions to write an equation for the line in point-slope form and general form.

Mathematics

1 answer:

Tatiana [17]3 years ago

5 0

Answer:

y=4/7 x +99/7 general form

y2-y1=m(x2-x1) =y2-9=4/7(x-(-9)

Step-by-step explanation:

4x - 7y - 2

slope=4/7 and b=-2/7

y=4/7 x-2/7

the slopes of parallel lines are equal. Note that two lines are parallel if their slopes are equal and they have different y-intercepts

point slope form : y2-y1=m(x2-x1) =y2-9=m(x-(-9)

y-9=4/7(x+9)

y=4/7x+36/7+9

y=4/7 x +99/7 general form

I hope it works

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Multiple-Choice Integration, Picture Included, Please Include Work

ValentinkaMS [17]

$\displaystyle\int_0^2\sqrt{4-x^2}\,\mathrm dx$

Recall that a circle of radius 2 centered at the origin has equation

$x^2+y^2=4\implies y=\pm\sqrt{4-x^2}$

where the positive root gives the top half of the circle in the x-y plane. The definite integral corresponds to the area of the right half of this top half. Since the area of a circle with radius $r$ is $\pi r^2$ , it follows that the area of a quarter-circle would be $\dfrac{\pi r^2}4$ .

You have $r=2$ , so the definite integral is equal to $\dfrac{2^2\pi}4=\pi$ .

Another way to verify this is to actually compute the integral. Let $x=2\sin u$ , so that $\mathrm dx=2\cos u\,\mathrm du$ . Now

$\displaystyle\int_0^2\sqrt{4-x^2}\,\mathrm dx=\int_0^{\pi/2}\sqrt{4-(2\sin u)^2}(2\cos u)\,\mathrm du=4\int_0^{\pi/2}\cos^2u\,\mathrm du$

Recall the half-angle identity for cosine:

$\cos^2u=\dfrac{1+\cos2u}2$

This means the integral is equivalent to

$\displaystyle2\int_0^{\pi/2}(1+\cos 2u)\,\mathrm du=2u+\sin2u\bigg|_{u=0}^{u=\pi/2}=\pi$