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3 years ago

What is the area of a circle?

1 answer:

8 0

The area of a circle is the size of the 2-dimensional space inside the circle's
closed curved boundary.

The area can be calculated in terms of known linear measurements of the circle:

-- Area = (π) x (radius)²

-- Area = (π/4) x (diameter)²

-- Area = (1/2) x (circumference) x (radius)

-- Area = (1/4) x (circumference) x (diameter)

Any of these formulas will give you the area. The one you decide to use
just depends on what you already know about the circle.

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Step-by-step explanation:

Hey There!

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3 years ago

Read 2 more answers

Explain the steps to finding the vertex of g(x) = 3x2 + 12x + 15

RUDIKE [14]

Answer:

The vertex of this parabola, $(-2, 3)$ , can be found by completing the square.

Step-by-step explanation:

The goal is to express this parabola in its vertex form:

$g(x) = a\, (x - h)^2 + k$ ,

where $a$ , $h$ , and $k$ are constants. Once these three constants were found, it can be concluded that the vertex of this parabola is at $(h,\, k)$ .

The vertex form can be expanded to obtain:

$\begin{aligned}g(x)&= a\, (x - h)^2 + k \\ &= a\, \left(x^2 - 2\, x\, h + h^2\right) + k = a\, x^2 - 2\, a\, h\, x + \left(a\,h^2 + k\right)\end{aligned}$ .

Compare that expression with the given equation of this parabola. The constant term, the coefficient for $x$ , and the coefficient for $x^2$ should all match accordingly. That is:

$\left\lbrace\begin{aligned}& a = 3 \\ & -2\,a\, h = 12 \\& a\, h^2 + k = 15\end{aligned}\right.$ .

The first equation implies that $a$ is equal to $3$ . Hence, replace the " $a\!$ " in the second equation with $3\!$ to eliminate $\! a$ :

$(-2\times 3)\, h = 12$ .

$h = -2$ .

Similarly, replace the " $a$ " and the " $h$ " in the third equation with $3$ and $(-2)$ , respectively:

$3 \times (-2)^2 + k = 15$ .

$k = 3$ .

Therefore, $g(x) = 3\, x^2 + 12\, x + 15$ would be equivalent to $g(x) = 3\, (x - (-2))^2 + 3$ . The vertex of this parabola would thus be:

$\begin{aligned}&(-2, \, 3)\\ &\phantom{(}\uparrow \phantom{,\,} \uparrow \phantom{)} \\ &\phantom{(}\; h \phantom{,\,} \;\;k\end{aligned}$ .

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