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3 years ago

Find the center-radius form of the equation of the circle described.

Mathematics

1 answer:

Kitty [74]3 years ago

8 0

Answer:

Correct answer: (x-√2)² + (y-√5)² = 3

Step-by-step explanation:

Given data: Center (x,y) = (√2,√5) and r = √3

The canonical or cartesian form of the equation of the circle is:

( x-p )² + ( y-q )² = r²

Where p is the x coordinate of the center, q is the y coordinate of the center and r is the radius of the circle.

God is with you!!!

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4 years ago

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Angelina_Jolie [31]

Answer:

See explanation

Step-by-step explanation:

A. Gasoline fill-up fee = $4.50

Cost per hour = $40

Average cost = C

Number of hours the scootcar is rented = h

Cost per h hours = $40h

Total cost = $40h + $4.50

Hence,

C = 40h+4.5

B. To plot the graph of the function, find h- and C- intercepts.

When h = 0, then C = 4.5, and we have point (0,4.5)

When C = 0, then 0 = 40h + 4.5, h = -45/400 = -0.1125, and we have point (-0.1125,0)

Plot these two points and connect them with a straight line.

Graph is attached, but you should take only that part of the graph that corresponds to points with h > 0, because the number of hours cannot be negative

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4 years ago

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aleksandr82 [10.1K]

This is one pathway to prove the identity.

Part 1

$\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{1}{\tan(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\cot(\theta) = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)*\sin(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)(1-\cos(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 2

$\frac{\sin^2(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)-\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-(\cos(\theta)-\cos^2(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-\cos(\theta)+\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 3

$\frac{\sin^2(\theta)+\cos^2(\theta)-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1}{\sin(\theta)} = \frac{1}{\sin(\theta)} \ \ {\checkmark}\\\\$

As the steps above show, the goal is to get both sides be the same identical expression. You should only work with one side to transform it into the other. In this case, the left side transforms while the right side stays fixed the entire time. The general rule is that you should convert the more complicated expression into a simpler form.

We use other previously established or proven trig identities to work through the steps. For example, I used the pythagorean identity $\sin^2(\theta)+\cos^2(\theta) = 1$ in the second to last step. I broke the steps into three parts to hopefully make it more manageable.