What is the standard form of the equationCenter (-14,5), circumference: 22π

Mathematics

1 answer:

IRISSAK [1]2 years ago

6 0

The standard equation for a circle with center at (h,k) and radius r is

(x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center of the circle.

The formula for the circumference of a circle is C = 2pi*r. In this particular problem, we need to determine the radius of the circle. That radius is: r = C/[2pi]. Here, C = 22pi, so we get r = 22pi/[2pi], and so r^2 = 11^2.

Putting to use the given info, we have:

(x+14)^2 + (y-5)^2 = 11^2

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3 years ago

! + 3 − 5 = 0; by completing the square

mylen [45]

Answer:

x = 1.19258 and x = −4.19258

Step-by-step explanation:

The given equation is :

$x^2+3x-5=0$

Keep x terms on the left and move the constant to the right side by adding it on both sides.

$x^2+3x=5$

Take half of the x term and square it.

$(3\times \dfrac{1}{2})^2=\dfrac{9}{4}$

then add the result to both sides

$x^2+3x+\dfrac{9}{4}=5+\dfrac{9}{4}\\\\(x+\dfrac{3}{2})^2=\dfrac{29}{4}\\\\(x+\dfrac{3}{2})=\sqrt{\dfrac{29}{4}}\\\\x=\dfrac{-3}{2}+\sqrt{\dfrac{29}{4}}\\\\x=\dfrac{-3}{2}-\sqrt{\dfrac{29}{4}}$