Question

find the center and radius of the circle.
$x^2 + y^2 + 4x + 2y - 76 = 0$
\\(\bigcirc\\) center: (-1, -2); radius: 9
\\(\bigcirc\\) center: (2, 1); radius: 81
\\(\bigcirc\\) center: (1, 2); radius: 81
\\(\bigcirc\\) center: (-2, -1); radius: 9
question 20
0.1 pts
find the function value. if the result is irrational, round your answer to the nearest thousandth.
let \\(f(x) = 6^x\\). find \\(f(-3)\\).
\\(\bigcirc\\) -18
\\(\bigcirc\\) \\(\frac{1}{729}\\)
\\(\bigcirc\\) \\(\frac{1}{216}\\)
\\(\bigcirc\\) \\(-\frac{1}{18}\\)

Explanation:

First Question (Circle Equation)

Step1: Rewrite the circle equation

The general form of a circle equation is \(x^{2}+y^{2}+Dx + Ey+F = 0\), and the standard form is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius. For the equation \(x^{2}+y^{2}+4x + 2y-76 = 0\), we complete the square for \(x\) and \(y\) terms.
For \(x\) terms: \(x^{2}+4x=(x + 2)^{2}-4\)
For \(y\) terms: \(y^{2}+2y=(y + 1)^{2}-1\)
Substitute these into the original equation: \((x + 2)^{2}-4+(y + 1)^{2}-1-76 = 0\)

Step2: Simplify to standard form

Simplify the equation: \((x + 2)^{2}+(y + 1)^{2}=4 + 1+76=81\)
Which is \((x + 2)^{2}+(y + 1)^{2}=9^{2}\). So the center is \((-2,-1)\) and radius is \(9\).

Second Question (Function Value)

Step1: Substitute \(x=-3\) into \(f(x)\)

Given \(f(x)=6^{x}\), substitute \(x = - 3\) into the function: \(f(-3)=6^{-3}\)

Step2: Use exponent rule

Recall the exponent rule \(a^{-n}=\frac{1}{a^{n}}\), so \(6^{-3}=\frac{1}{6^{3}}\)

Step3: Calculate \(6^{3}\)

\(6^{3}=6\times6\times6 = 216\), so \(f(-3)=\frac{1}{216}\)

Answer:

center: \((-2, -1)\); radius: \(9\)