I need the answer to the below question

Find the equation of a circle with a center at (7,2) and a point on the circle at (2,5)?

monitta

Answer:

$(x-7)^2+(y-2)^2=34$

Step-by-step explanation:

We want to find the equation of a circle with a center at (7, 2) and a point on the circle at (2, 5).

First, recall that the equation of a circle is given by:

$(x-h)^2+(y-k)^2=r^2$

Where (<em>h, k</em>) is the center and <em>r</em> is the radius.

Since our center is at (7, 2), <em>h</em> = 7 and <em>k</em> = 2. Substitute:

$(x-7)^2+(y-2)^2=r^2$

Next, the since a point on the circle is (2, 5), <em>y</em> = 5 when <em>x</em> = 2. Substitute:

$(2-7)^2+(5-2)^2=r^2$

Solve for <em>r: </em>

Simplify. Thus:

$25+9=r^2$

Finally, add:

$r^2=34$

We don't need to take the square root of both sides, as we will have the square it again anyways.

Therefore, our equation is:

$(x-7)^2+(y-2)^2=34$

4 0

2 years ago

ASAP PLEASE!!!

mario62 [17]

It should be D Vietnam I believe and by the way you have this under mathematics.

8 0

3 years ago

Pls help me for once....

ryzh [129]

The answers are C and also D

3 0

3 years ago

Help me with this...

marin [14]

i. 171

ii. 162

iii. 297

Solution,

n(U)= 630

n(I)= 333

n(T)= 168

i. Let n(I intersection T ) be X

$333 - x + x + 468 - x = 630 \\ or \: 333 + 468 - x = 630 \\ or \: 801 - x = 630 \\ or \: - x = 630 - 801 \\ or \: - x = - 171 \\ x = 171$

<h3>ii.n(only I)= n(I) - n(I intersection T)</h3><h3> = 333 - 171</h3><h3> = 162</h3>

<h3>iii. n ( only T)= n( T) - n( I intersection T)</h3><h3> = 468 - 171</h3><h3> = 297</h3>

<h3>Venn- diagram is shown in the attached picture.</h3>

Hope this helps...

Good luck on your assignment...

4 0

3 years ago

Water pours into a tank at the rate of 2000 cm3/min. The tank is cylindrical with radius 2 meters. How fast is the height of wat

Gennadij [26K]

Volume of water in the tank:

$V=\pi (2\,\mathrm m)^2h=\pi(200\,\mathrm{cm})^2h$

Differentiate both sides with respect to time <em>t</em> :

$\dfrac{\mathrm dV}{\mathrm dt}=\pi(200\,\mathrm{cm})^2\dfrac{\mathrm dh}{\mathrm dt}$

<em>V</em> changes at a rate of 2000 cc/min (cubic cm per minute); use this to solve for d<em>h</em>/d<em>t</em> :

$2000\dfrac{\mathrm{cm}^3}{\rm min}=\pi(40,000\,\mathrm{cm}^2)\dfrac{\mathrm dh}{\mathrm dt}$

$\dfrac{\mathrm dh}{\mathrm dt}=\dfrac{2000}{40,000\pi}\dfrac{\rm cm}{\rm min}=\dfrac1{20\pi}\dfrac{\rm cm}{\rm min}$

(The question asks how the height changes at the exact moment the height is 50 cm, but this info is a red herring because the rate of change is constant.)

7 0

3 years ago