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3 years ago

What is the center of a circle? What is the radius of the circle?

Mathematics

2 answers:

lutik1710 [3]3 years ago

6 0

Answer:

Center: (-9, -7)

Radius: 5

Step-by-step explanation:

erik [133]3 years ago

6 0

The center would be (-9,-7) the radius would be 5
i did something like this on edunity

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9x+3y=18 whats is the equation when solved for y

lidiya [134]

Answer:

y=6-3x

Step-by-step explanation:

first subtract 9x from 9x+3y=18

-9x -9x

so you get 3y=18-9x or 3y=-9x+18 then divide both sides by 3

so you get y=6-3x hope this helps :)

6 0

3 years ago

Andrea had 32 problems to do for mathematics homework over the weekend. She did 1/4 of the problems on Friday night, and on Satu

igor_vitrenko [27]

Answer:

Step-by-step explanation:

Problems at start = 32

-Problems done Friday = ¼ × 32 = 8

Problems still to be done = 24

Andrea had to do 24 problems on Saturday to complete the homework.

6 0

3 years ago

7 2 2 2 3 8 0 58 9 3 7 10 0

worty [1.4K]

Answer:

Wut...?

Step-by-step explanation:

5 0

3 years ago

Convert 9.3 kribs into grees given that 4.2 kribs = 18 grees.

maria [59]

Answer:

39.9 grees

Step-by-step explanation:

Since 4.2 kribs = 18 grees

Then,

1 krib = 18/4.2 grees

9.3 kribs = (18/4.2) x 9 3

9.3 kribs = 39.9 grees.

4 0

3 years ago

Integrate the following

enot [183]

I suppose you mean to have the entire numerator under the square root?

$\displaystyle\int_2^4\frac{\sqrt{x^2-4}}{x^2}\,\mathrm dx$

We can use a trigonometric substitution to start:

$x=2\sec t\implies\mathrm dx=2\sec t\tan t\,\mathrm dt$

Then for $x=2$ , $t=\sec^{-1}1=0$ ; for $x=4$ , $t=\sec^{-1}2=\frac\pi3$ . So the integral is equivalent to

$\displaystyle\int_0^{\pi/3}\frac{\sqrt{(2\sec t)^2-4}}{(2\sec t)^2}(2\sec t\tan t)\,\mathrm dt=\int_0^{\pi/3}\frac{\tan^2t}{\sec t}\,\mathrm dt$

We can write

$\dfrac{\tan^2t}{\sec t}=\dfrac{\frac{\sin^2t}{\cos^2t}}{\frac1{\cos t}}=\dfrac{\sin^2t}{\cos t}=\dfrac{1-\cos^2t}{\cos t}=\sec t-\cos t$

so the integral becomes

$\displaystyle\int_0^{\pi/3}(\sec t-\cos t)\,\mathrm dt=\boxed{\ln(2+\sqrt3)-\frac{\sqrt3}2}$

7 0

3 years ago