All categories

3 years ago

What are the center and radius of the circle given by the equation ?

Mathematics

1 answer:

valina [46]3 years ago

3 0

General formula of the circle

(x-a)² + (y - b)² = r²,

where (a,b) - coordinates of the center, r - radius of the circle.

(x-3)²+(y+4)²=4

(x-3)²+(y+4)²=2²

So, the center is (3, -4) and radius r = 2.

A. (3, –4); 2

You might be interested in

The vertex of the parabola below is at the point (2, 4), and the point (3, 6) is on the parabola. What is the equation of the pa

Liula [17]

<span>C and B are *candidates* for being the correct solution
because f(2) = 4 for both.
</span>
A is the correct solution because f(3)=6

4 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csf%5Clim_%7Bx%20%5Cto%200%20%7D%20%5Cfrac%7B1%20-%20%5Cprod%20%5Climits_%

xxTIMURxx [149]

To demonstrate a method for computing the limit itself, let's pick a small value of n. If n = 3, then our limit is

$\displaystyle \lim_{x \to 0 } \frac{1 - \prod \limits_{k = 2}^{3} \sqrt[k]{\cos(kx)} }{ {x}^{2} }$

Let a = 1 and b the cosine product, and write them as

$\dfrac{a - b}{x^2}$

with

$b = \sqrt{\cos(2x)} \sqrt[3]{\cos(3x)} = \sqrt[6]{\cos^3(2x)} \sqrt[6]{\cos^2(3x)} = \left(\cos^3(2x) \cos^2(3x)\right)^{\frac16}$

Now we use the identity

$a^n-b^n = (a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots a^2b^{n-3}+ab^{n-2}+b^{n-1}\right)$

to rationalize the numerator. This gives

$\displaystyle \frac{a^6-b^6}{x^2 \left(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5\right)}$

As x approaches 0, both a and b approach 1, so the polynomial in a and b in the denominator approaches 6, and our original limit reduces to

$\displaystyle \frac16 \lim_{x\to0} \frac{1-\cos^3(2x)\cos^2(3x)}{x^2}$

For the remaining limit, use the Taylor expansion for cos(x) :

$\cos(x) = 1 - \dfrac{x^2}2 + \mathcal{O}(x^4)$

where $\mathcal{O}(x^4)$ essentially means that all the other terms in the expansion grow as quickly as or faster than x⁴; in other words, the expansion behaves asymptotically like x⁴. As x approaches 0, all these terms go to 0 as well.

Then

$\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 2x^2\right)^3 \left(1 - \frac{9x^2}2\right)^2$

$\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 6x^2 + 12x^4 - 8x^6\right) \left(1 - 9x^2 + \frac{81x^4}4\right)$

$\displaystyle \cos^3(2x) \cos^2(3x) = 1 - 15x^2 + \mathcal{O}(x^4)$

so in our limit, the constant terms cancel, and the asymptotic terms go to 0, and we end up with

$\displaystyle \frac16 \lim_{x\to0} \frac{15x^2}{x^2} = \frac{15}6 = \frac52$

Unfortunately, this doesn't agree with the limit we want, so n ≠ 3. But you can try applying this method for larger n, or computing a more general result.

Edit: some scratch work suggests the limit is 10 for n = 6.

6 0

2 years ago

At a zoo, the lion pen has a ring-shaped sidewalk around it. The outer edge of the sidewalk is a circle with a radius of 12 m. T

cestrela7 [59]

Answer:

The answer would be 251.2 for the side walk if the radius for the whole circle is 12 and the small circle's radius would be 8. Only for the answer with the 8 m and 12 m radius NOT the 11m and 9 m

The equation would be:

144 x 3.14 - 64 x 3.14 = 251.2

Step-by-step explanation:

First, you need to find out what the whole circle is:

The whole circle is 452.16

Then, find the inner circle :

8 x 8 x 3.14

200.96

Last, subtract the whole by the inner:

452.16 - 200.96 = 251.2

hope this helps!

5 0

3 years ago

What the cube root between 100 and 600

gizmo_the_mogwai [7]

A list of cube roots between 100 and 600 are
5x5x5= 125
6x6x6=216
7x7x7=343
8x8x8=512

8 0

2 years ago

Joey Chestnut is trying to break his own record for eating the most whole hot dogs in 10 minutes. He needs to eat more than 72 h

anastassius [24]

The picture below is the answer to your question. I solved it myself when I was doing my homework.

5 0

3 years ago

Read 2 more answers