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3 years ago

11

Write the equation of the circle with center (−3, −2) and (4, 5) a point on the circle.

2 answers:

vivado [14]3 years ago

8 0

Http://www.personal.kent.edu/~bosikiew/Algebra-handouts/circles.pdf

I found this is may help but I don't know for sure.

masha68 [24]3 years ago

4 0

Answer:

B) (x + 3)2 + (y + 2)2 = 98

Step-by-step explanation:

You might be interested in

Convert to equation: four times a number increased by 8 is 24

creativ13 [48]

Answer:

4*4

Step-by-step explanation:

since 4 times 4 is 16 then if you add 8 its 24

4 0

3 years ago

Y equals -2 over 5 x + 1 y-intercept

kipiarov [429]

Y = -2/5x + 1

I think your question is what is the y-intercept? It would be 1. -2/5 is the slope.

4 0

3 years ago

Find all solutions in the interval [0, 2π). <br> 2 sin2x = sin x

tester [92]

Answer:

The solutions on the given interval are :

$0$

$\pi$

$\cos^{-1}(\frac{1}{4})$

$-\cos^{-1}(\frac{1}{4})+2\pi$

Step-by-step explanation:

We will need the double angle identity $\sin(2x)=2\sin(x)\cos(x)$ .

Let's begin:

$2\sin(2x)=\sin(x)$

Use double angle identity mentioned on left hand side:

$2\cdot 2\sin(x)\cos(x)=\sin(x)$

Simplify a little bit on left side:

$4\sin(x)\cos(x)=\sin(x)$

Subtract $\sin(x)$ on both sides:

$4\sin(x)\cos(x)-\sin(x)=0$

Factor left hand side:

$\sin(x)[4\cos(x)-1]=0$

Set both factors equal to 0 because at least of them has to be 0 in order for the equation to be true:

$\sin(x)=0 \text{ or } 4\cos(x)-1=0$

The first is easy what angles $\theta$ are $y$ -coordinates on the unit circle 0. That happens at $0$ and $\pi$ on the given range of $x$ (this $x$ is not be confused with the $x$ -coordinate).

Now let's look at the second equation:

$4\cos(x)-1=0$

Isolate $\cos(x)$ .

Add 1 on both sides:

$4 \cos(x)=1$

Divide both sides by 4:

$\cos(x)=\frac{1}{4}$

This is not as easy as finding on the unit circle.

We know $\arccos( )$ will render us a value between $0$ and $2\pi$ .

So one solution on the given interval for x is $x=\cos^{-1}(\frac{1}{4})$ .

We know cosine function is even.

So an equivalent equation is:

$\cos(-x)=\frac{1}{4}$

Apply $\cos^{-1}$ to both sides:

$-x=\cos^{-1}(\frac{1}{4})$

Multiply both sides by -1:

$x=-\cos^{-1}(\frac{1}{4})$

This going to be negative in the 4th quadrant but if we wrap around the unit circle, $2\pi$ , we will get an answer between $0$ and $2\pi$ .

So the solutions on the given interval are :

$0$

$\pi$

$\cos^{-1}(\frac{1}{4})$

$-\cos^{-1}(\frac{1}{4})+2\pi$

5 0

3 years ago

What’s the equation of the horizontal asymptote

Maksim231197 [3]

By inspection, the equation for the horizontal asymptote is y = -3.

6 0

3 years ago

NEED HELP ASAP!!! PLZ HELP!!!

anyanavicka [17]

just guess

jk i think the last one

4 0

3 years ago