If you were to take a cross section of a right circular cone parallel to its base, what shape would you get?

Mathematics

1 answer:

Leya [2.2K]2 years ago

3 0

The cross section of a right circular cone parallel to its base is a circle.

<h3>What is a three dimensional shape?</h3>

A three dimensional shape is a shape that has length, width and height. Examples of three dimensional shapes are cone, cylinder, prism and pyramid.

A two dimensional shape is a shape that have both length and width. Examples of two dimensional shape are circle, rectangle, square.

A cone is a three-dimensional solid geometric shape having a circular base and a pointed edge at the top

The cross section of a right circular cone parallel to its base is a circle.

Find out more on three dimensional shape at: brainly.com/question/6636176

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Identify the x-intercepts of the function below f(x)=x^2+12x+24

damaskus [11]

ANSWER:

x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

SOLUTION:

Given, $f(x)=x^{2}+12 x+24$ -- eqn 1

x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.

Now, let us find the zeroes using quadratic formula for f(x) = 0.

$X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Here, for (1) a = 1, b= 12 and c = 24

$X=\frac{-(12) \pm \sqrt{(12)^{2}-4 \times 1 \times 24}}{2 \times 1}$

$\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}$