All categories

3 years ago

What is the equation of a circle centered at (-3,1) with radius 7?

1 answer:

8 0

The standard form of a circle is $(x-h)^2+(y-k)^2=r^2$ , where h and k are the coordinates of the center and r is the radius. We will fill in accordingly with what we have. $(x-(-3))^2+(y-1)^2=7^2$ . Simplifying a bit we have $(x+3)^2+(y-1)^2=49$ . That's the equation for your circle.

You might be interested in

HELP NEEDED ASAP GIVING BRAINLIEST AND PLEASE EXPLAIN

Dennis_Churaev [7]

Answer: (B is the answer.

Step-by-step explanation: The graph I attached corresponds with B.

Cross -3 by -2.

5 0

3 years ago

Sarah has 25 candies. Sarah gave 2/5 of the candies to her sister. Of the amount left she gave 2/3 to her friend. How many candi

satela [25.4K]

Answer:

5 candies

Step-by-step explanation:

Total of candy Sarah has = 25candies

If Sarah gave 2/5 of the candies to her sister, the total amount she gave her sister will be 2/5 of 25

2/5 of 25 = 10

The remaining candy left will be 25-10 which is 15.

If she gave 2/3 of amount left to her friend, the amount she gave her friend will be 2/3×15 = 10candies.

The amount of candy Sarah have left will be 15- 10 which is 5candies.

3 0

2 years ago

Marine biked 3/4 of a mile on Monday. On Tuesday he biked 5 times this distance. The number of miles marine bikes on Tuesday is

PolarNik [594]

It's simple. First, multiply 3/4 by 5. As a decimal, this equals 3.75. Therefore, the number of miles marine bikes on Tuesday is between 3 and 4.

Hope I helped! :)

4 0

2 years ago

The lowest recorded temperature in a town was 4ºF, without the windchill. With the wind, the temperature felt like 10ºF below ze

deff fn [24]

Answer:

umm it think it is 14

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

The equation giving a family of ellipsoids is u = (x^2)/(a^2) + (y^2)/(b^2) + (z^2)/(c^2) . Find the unit vector normal to each

Fynjy0 [20]

Answer:

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Step-by-step explanation:

Given equation of ellipsoids,

$u\ =\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}$

The vector normal to the given equation of ellipsoid will be given by

$\vec{n}\ =\textrm{gradient of u}$

$=\bigtriangledown u$

$=\ (\dfrac{\partial{}}{\partial{x}}\hat{i}+ \dfrac{\partial{}}{\partial{y}}\hat{j}+ \dfrac{\partial{}}{\partial{z}}\hat{k})(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2})$

$=\ \dfrac{\partial{(\dfrac{x^2}{a^2})}}{\partial{x}}\hat{i}+\dfrac{\partial{(\dfrac{y^2}{b^2})}}{\partial{y}}\hat{j}+\dfrac{\partial{(\dfrac{z^2}{c^2})}}{\partial{z}}\hat{k}$

$=\ \dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}$

Hence, the unit normal vector can be given by,

$\hat{n}\ =\ \dfrac{\vec{n}}{\left|\vec{n}\right|}$

$=\ \dfrac{\dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}}{\sqrt{(\dfrac{2x}{a^2})^2+(\dfrac{2y}{b^2})^2+(\dfrac{2z}{c^2})^2}}$

$=\ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Hence, the unit vector normal to each point of the given ellipsoid surface is

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

3 0

3 years ago