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Gre4nikov [31]
2 years ago
12

A circle has its center at (0,0) and passes through the point (0,9). What is the standard equation of the circle?

Mathematics
1 answer:
Alexxandr [17]2 years ago
7 0

Answer:

x^2+y^2=9^2

Step-by-step explanation:

The standard equation of a circle is:

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

Where the center of a circle is (h,k) and r is the radius of the circle.  In this case because a circle is equidistant from the center and we have a point where it passes through 9 that means that the radius is 9.  However, since the standard equation states that we must write r in the form of r^2 this means that r^2=9^2.  Therefore by plugging in the values we have:

(x-0)^2+(y-0)^2=9^2\\\\x^2+y^2=9^2

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Find f(x) and g(x) so that the function can be described as y = f(g(x)).<br><br> y = 4/x^2+9
dlinn [17]

I'm assuming all of (x^2+9) is in the denominator. If that assumption is correct, then,

One possible answer is f(x) = \frac{4}{x},  \ \ g(x) = x^2+9

Another possible answer is f(x) = \frac{4}{x+9}, \ \ g(x) = x^2

There are many ways to do this. The idea is that when we have f( g(x) ), we basically replace every x in f(x) with g(x)

So in the first example above, we would have

f(x) = \frac{4}{x}\\\\f( g(x) ) = \frac{4}{g(x)}\\\\f( g(x) ) = \frac{4}{x^2+9}

In that third step, g(x) was replaced with x^2+9 since g(x) = x^2+9.

Similar steps will happen with the second example as well (when g(x) = x^2)

4 0
3 years ago
an isosceles triangle has congruent sides of 20cm. the base is 10cm. find the height of the triangle.​
dolphi86 [110]

Answer:

\large\boxed{A_\triangle=25\sqrt{15}\ cm^2}

Step-by-step explanation:

Look at the picture.

The formula of an area of a triangle:

A_\triangle=\dfrac{bh}{2}

<em>b</em><em> - base</em>

<em>h</em><em> - height</em>

<em />

We need a length of a height.

Use the Pythagorean theorem:

leg^2+leg^2=hypotenuse^2

We have:

leg=5,\ leg=h,\ hypotenuse=20

Substitute:

5^2+h^2=20^2

25+h^2=400             <em>subtract 25 from both sides</em>

h^2=375\to h=\sqrt{375}\\\\h=\sqrt{(25)(15)}\\\\h=\sqrt{25}\cdot\sqrt{15}\\\\h=5\sqrt{15}\ cm

Calculate the area:

A_\triangle=\dfrac{(10)(5\sqrt{15})}{2}=\dfrac{50\sqrt{15}}{2}=25\sqrt{15}\ cm^2

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3 years ago
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The answer is 24. You subtract 40 - 16.

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Answer:

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Step-by-step explanation:

You use the equation length times width to find the area. 15 x 8 = 120. This equation works for any rectangle

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