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2 years ago

Please help me with this

Mathematics

2 answers:

Stella [2.4K]2 years ago

3 0

I would but I can’t see the pic.

vodomira [7]2 years ago

3 0

I would but I can’t see anything there bud

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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 (h, k) is the center and r is the radius.

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

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

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

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

Solve for r:

$(-5)^2+(3)^2=r^2$

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:

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

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3 years ago

Solve the following system by graphing.

Masja [62]

Hello :
the solution is : (3, -1)
put x= 3 and y = -1
you have : 3-(-1) = 4 and 3+(-1) = 2 ..... right

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3 years ago

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Can someone give me some tips of how to study. I have already tried writing everything in my lesson but it doesn't seem to work

Ainat [17]

Study all notes, reread the chapters again. Have someone ask questions on the chapters page by page. This always has worked for me. Plus try to do this again the night before the test. You will be surprised how much you can remember by doing it again the night before the test. Hope this helps.

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