Find the equation of the circle that satisfies the given conditions. Radius 2 and center (4, -7)?

1 answer:

6 0

$The\ equation\ of\ the\ circle:\\\\(x-a)^2+(y-b)^2=r^2\\\\where\\\\center:(a;\ b)\\\\radius:r$

$Center:(4;-7);\ radius:r=2\\\\(x-4)^2+(y+7)^2=2^2\\\\(x-4)^2+(y+7)^2=4$

You might be interested in

Write the distance in scientific notation MATH HELP

Stella [2.4K]

58000000=5.8⋅10

The 10 is to the 7th power.

I hope that helps!!!!!!

: )

4 0

3 years ago

Read 2 more answers

Determine the quotient, q(x), and remainder, r(x) when

Vinvika [58]

ANSWER

Quotient:

$q(x) = 4 {x}^{2} + 3x + 2$

Remainder:

$r(x) = 3x + 1$

EXPLANATION

The given functions are

$f(x) = 12 {x}^{4} + 21 {x}^{3} + 31 {x}^{2} + 21x + 9$

and

$g(x) = 3 {x}^{2} + 3x + 4$

We want to find the remainder and quotient when f(x) is divided by g(x).

We perform the long division as shown in the attachment.

The quotient is

$q(x) = 4 {x}^{2} + 3x + 2$

The remainder is

$r(x) = 3x + 1$

5 0

3 years ago

A. 6x^3+18x b.6x^2+3 c.36x^2+18 d. 36x^2+3

zzz [600]

The answer: 6x^3+18x

The Explanation: You need to uses distributive property to 6x to x^2+3 which will be 6x^3+18x

6 0

3 years ago

What is the sum of the geometric series in which a1 = −5, r = 3, and an = −405?

xxMikexx [17]

$a_1=-5;\ r=3;\ a_n=-405\\\\a_n=a_1r^{n-1}\\\\subtitute\\\\-5\cdot3^{n-1}=-405\ \ \ |divide\ both\ sides\ by\ (-5)\\\\3^{n-1}=81\\\\3^{n-1}=3^4\iff n-1=4\to n=5\\\\S_n=\dfrac{a_1(1-r^n)}{1-r}\\\\S_5=\dfrac{-5(1-3^5)}{1-3}=\dfrac{-5(1-243)}{-2}=\dfrac{5\cdot(-242)}{2}=5\cdot(-121)=-605\\\\\\Answer:\boxed{S_5=-605}$

4 0

3 years ago

What is 20% by 37 decreased

Ghella [55]

Answer:

37 decreased by 20% is 29.6.

Step-by-step explanation:

To do this, what we do is simply take 20% of 37 and subtract it from 37.

20% of 37. A trick to easily figure this out is multipling 37 by 20 and dividing by 100.

37*20 = 740

740 / 100 = 7.40

37 - 7.40 = 29.6

37 decreased by 20% is 29.6.

6 0

2 years ago

Read 2 more answers