What is the standard form of equation of a circle with center at (-2, -3) and passing through the point (-2, 0)?

1 answer:

3 0

Hello :
an equation of the circle Center at the A(a,b) and ridus : r is :
(x-a)² +(y-b)² = r²
in this exercice : a = -2 and b =-3 (Center A(-2,-3))
r = AP......(P(-2,0))
r² = (AP)²
r² = (-2+2)² +(-3-0)² = 9
an equation of the circle that satisfies the stated conditions.
Center A(-2,-3) , passing through P(-2, 0) is : (x+2)² +(y+3)² = 9

You might be interested in

As the sample size becomes larger, the sampling distribution of the sample mean approaches a _____. a. binomial distribution b.

kkurt [141]

Answer:

c. normal probability distribution

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

In this question:

By the Central Limit Theorem, it is a normal distribution, so option c.

7 0

2 years ago

What is the remainder when 18,049 is divided by 14? (Input numeric values only)

ioda

Answer:

18049/14= 1289 with a remainder of 3

3 0

3 years ago

Read 2 more answers

Calculus: Help ASAP

Serggg [28]

$\bf \displaystyle \int\limits_{-1}^{0}~(4x^6+2x)^3(12x^5+1)\cdot dx\\\\
-------------------------------\\\\
u=4x^6+2x\implies \cfrac{du}{dx}=24x^5+2\implies \cfrac{du}{2(12x^5+1)}=dx\\\\
-------------------------------\\\\
\displaystyle \int\limits_{-1}^{0}~u^3\underline{(12x^5+1)}\cdot\cfrac{du}{2\underline{(12x^5+1)}}\implies \cfrac{1}{2}\int\limits_{-1}^{0}~u^3\cdot du\\\\
-------------------------------\\\\$

$\bf \textit{now, we'll change the bounds, using u(x)}
\\\\\\
u(-1)=4(-1)^6+2(-1)\implies u(-1)=2
\\\\\\
u(0)=4(0)^6+2()\implies u(0)=0\\\\
-------------------------------\\\\
\displaystyle \cfrac{1}{2}\int\limits_{2}^{0}~u^3\cdot du\implies \left. \cfrac{1}{2}\cdot \cfrac{u^4}{4} \right]_{2}^{0}\implies \left. \cfrac{u^4}{8} \right]_{2}^{0}\implies [0]-[2]\implies -2$

5 0

3 years ago

Can somebody step by step explain how to solve this problem?? (3x^2y)(-7xy^3)

charle [14.2K]

Answer:

$\boxed{\bold{-21x^3y^4}}$