All categories

4 years ago

Write the standard equation of a circle with center (0, 3) passing through the point (2, 5)

Mathematics

1 answer:

marysya [2.9K]4 years ago

3 0

The equation is x² + (y-3)²=8

You might be interested in

D = 8.3 mm<br> Calculate the radius of the circle

EastWind [94]

Answer:

r= d/2

8.3/2= 4.15

Step-by-step explanation:

4 0

3 years ago

Write an equation in slope-intercept form of the line that passes through (0,6) and (3, -3).

snow_lady [41]

Answer:

y=-3x+6

Step-by-step explanation:

(y-y1)=m(x-x1)

m=(y2-y1)/(x2-x1)

m=(-3-6)/(3-0)

m=(-9)/(3)

m=-3

(choose whichever point for the first formula, i will be using(0,6))

(y-6)=-3(x-0)

y-6=-3x

y=-3x+6

8 0

3 years ago

Does anybody know the answer to this question?

Readme [11.4K]

Answer:

D, because each input has exactly one output.

5 0

2 years ago

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. Verify y

Drupady [299]

Answer:

$V = \frac{\pi^2}{8}$

$V = 1.23245$

Step-by-step explanation:

Given

$y = \cos 2x$

$y = 0; x = 0; x = \frac{\pi}{4}$

Required

Determine the volume of the solid generated

Using the disk method approach, we have:

$V = \pi \int\limits^a_b {R(x)^2} \, dx$

Where

$y = R(x) = \cos 2x$

$a = \frac{\pi}{4}; b =0$

So:

$V = \pi \int\limits^a_b {R(x)^2} \, dx$

Where

$y = R(x) = \cos 2x$

$a = \frac{\pi}{4}; b =0$

So:

$V = \pi \int\limits^a_b {R(x)^2} \, dx$

$V = \pi \int\limits^{\frac{\pi}{4}}_0 {(\cos 2x)^2} \, dx$

$V = \pi \int\limits^{\frac{\pi}{4}}_0 {\cos^2 (2x)} \, dx$

Apply the following half angle trigonometry identity;

$\cos^2(x) = \frac{1}{2}[1 + \cos(2x)]$

So, we have:

$\cos^2(2x) = \frac{1}{2}[1 + \cos(2*2x)]$

$\cos^2(2x) = \frac{1}{2}[1 + \cos(4x)]$

Open bracket

$\cos^2(2x) = \frac{1}{2} + \frac{1}{2}\cos(4x)$

So, we have:

$V = \pi \int\limits^{\frac{\pi}{4}}_0 {\cos^2 (2x)} \, dx$

$V = \pi \int\limits^{\frac{\pi}{4}}_0 {[\frac{1}{2} + \frac{1}{2}\cos(4x)]} \, dx$

Integrate

$V = \pi [\frac{x}{2} + \frac{1}{8}\sin(4x)]\limits^{\frac{\pi}{4}}_0$

Expand

$V = \pi ([\frac{\frac{\pi}{4}}{2} + \frac{1}{8}\sin(4*\frac{\pi}{4})] - [\frac{0}{2} + \frac{1}{8}\sin(4*0)])$

$V = \pi ([\frac{\frac{\pi}{4}}{2} + \frac{1}{8}\sin(4*\frac{\pi}{4})] - [0 + 0])$

$V = \pi ([\frac{\frac{\pi}{4}}{2} + \frac{1}{8}\sin(4*\frac{\pi}{4})])$

$V = \pi ([{\frac{\pi}{8} + \frac{1}{8}\sin(\pi)])$

$\sin \pi = 0$

So:

$V = \pi ([{\frac{\pi}{8} + \frac{1}{8}*0])$

$V = \pi *[{\frac{\pi}{8}]$

$V = \frac{\pi^2}{8}$

$V = \frac{3.14^2}{8}$

$V = 1.23245$

4 0

3 years ago

What is the answer for y=−16x^2 +119x +73

Zina [86]

Answer:

Im pretty sure the answer is y=375x+73

Im not really sure

6 0

3 years ago