Find an equation of the circle whose diameter has endpoints (-2, 3) and (-2, -6).

Mathematics

1 answer:

Pie3 years ago

3 0

Answer:

( -5,-5 )

Step-by-step explanation:

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Solve 54 - 10x s 20 + 7x.

konstantin123 [22]

9514 1404 393

Answer:

2 ≤ x

Step-by-step explanation:

Maybe you want to solve ...

54 -10x ≤ 20 +7x

34 ≤ 17x . . . . . . . . . . . . add 10x-20 to both sides

2 ≤ x . . . . . . . . . . . . . . divide by 17

6 0

2 years ago

A submarine descended 3/5 of a kilometer in 1/5 of an hour. What was the submarine’s rate of descent in kilometers per hour?

grandymaker [24]

Answer:

3 km per hour

Step-by-step explanation:

You can deduct the answer instinctively since both the distance and the time of the descent are expressed in /5... so you can just get rid of the '/5' in both instances to get the result: 3 kilometers in 1 hour.

But you can also find it easily using the cross-multiplication system:

$\frac{3}{5} km = \frac{1}{5} hour$

Then you multiply each side by 5 to get

3 km = 1 hour

Thus the rate of descent is of 3 km per hour.

3 0

3 years ago

Read 2 more answers

What's the area of this rectangle?

Bess [88]

well, the assumption is that is a rectangle, namely it has two equal pairs, so we can just find the length of one of the pairs to get the dimensions.

hmmmm let's say let's get the length of the segment at (-1,-3), (1,3) for its length

and

the length of the segment at (-1, -3), (-4, -2) for its width

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{length}{L}=\sqrt{[1-(-1)]^2+[3-(-3)]^2}\implies L=\sqrt{(1+1)^2+(3+3)^2} \\\\\\ L=\sqrt{4+36}\implies L=\sqrt{40} \\\\[-0.35em] ~\dotfill$

$\bf (\stackrel{x_1}{-1}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{-4}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{width}{w}=\sqrt{[-4-(-1)]^2+[-2-(-3)]^2}\implies w=\sqrt{(-4+1)^2+(-2+3)^2} \\\\\\ w=\sqrt{9+1}\implies w=\sqrt{10} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of the rectangle}}{A=Lw}\implies \sqrt{40}\cdot \sqrt{10}\implies \sqrt{400}\implies \boxed{20}$