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Alika [10]
2 years ago
12

(3+2a)² alguien sabe la respuesta?

Mathematics
1 answer:
asambeis [7]2 years ago
6 0

Answer:

3(3+2a)+2a(3+2a)

9+6a+6a+4a^2

9+12a+4a^2

ANS 4a^2+12a+9

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The graphs below have the same shape. What is the equation of the red graph?
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We claim that the answer is y=-x^2. Since the two graphs both open down, and all the answer choices, in addition to the equation of the blue graph, are quadratic polynomials, the leading coefficient must be negative. This immediately rules out answer choices A, B, and C, leaving D as the answer.

4 0
3 years ago
Find the equation of a circle that has a diameter with the endpoints given by the points A(-4,9) and B (- 2, - 3) )
Aleksandr-060686 [28]

The equation of the circle is (x+3)^{2}+(y-3)^{2}=37

Explanation:

Given that the endpoints of the circle are A(-4,9) and B(-2,-3)

We need to determine the equation of the circle.

<u>Center:</u>

The center of the circle can be determined using the midpoint formula,

Center=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})

Substituting the coordinates A(-4,9) and B(-2,-3), we get,

Center=(\frac{-4-2}{2},\frac{9-3}{2})

Center=(\frac{-6}{2},\frac{6}{2})

Center=(-3,3)

Thus, the center of the circle is (-3,3)

<u>Radius:</u>

The radius of the circle can be determined using the distance formula,

r=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}

Substituting the center (-3,3) and the endpoint (-4,9), we get,

r=\sqrt{\left(-4+3\right)^{2}+\left(9-3\right)^{2}}

r=\sqrt{\left(-1\right)^{2}+\left(6\right)^{2}}

r=\sqrt{1+36}

r=\sqrt{37}

Thus, the radius of the circle is \sqrt{37}

<u>Equation of the circle:</u>

The standard form of the equation of the circle is given by

(x-a)^{2}+(y-b)^{2}=r^{2}

where (a,b) is the center and r is the radius.

Substituting the values, we have,

(x+3)^{2}+(y-3)^{2}=(\sqrt{37})^{2}

(x+3)^{2}+(y-3)^{2}=37

Thus, the equation of the circle is (x+3)^{2}+(y-3)^{2}=37

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