All categories

3 years ago

What shapes can you obtain by taking a cross section of a cone?

Mathematics

1 answer:

SashulF [63]3 years ago

7 0

You can get a circle by cutting horizontally and a hyperbola by cutting vertically. You can also get an ellipse and a parabola by cutting at an angle depending on whether the cut comes out the other side or not.

You might be interested in

max is bisecting a segment. first. he places the compass on one endpoint and opens it to a width larger than half of the segment

spin [16.1K]

Answer:

Keep the compass the same width and place it on the other endpoint.

Step-by-step explanation:

Max is bisecting a segment. First, he places the compass on one endpoint and opens it to a width larger than half of the segment. Then he swings an arc on either side of the segment.

Next Max's step should be: keep the compass the same width and place it on the other endpoint.

He will draw the second arc of the same radius as the first one.

Two drawn arcs intersect at two points, Max will connect these points and get the segment which bisects the given one.

7 0

3 years ago

Read 2 more answers

The graphs of two rational functions of f and g are shown. One of them is given by the expression 2-3x/x. Which graph is it? Tha

Nesterboy [21]

Answer:

It's the left graph, y = f(x)

Step-by-step explanation:

If you find the x-intercepts, you'd solve (2-3x)/x = 0. This means you'd really solve 2-3x=0, which gives you x=3/2.

So the graph must have an x-intercept at (1.5, 0). Only f(x) has that.

8 0

3 years ago

Help me with trigonometry

poizon [28]

Answer:

See below

Step-by-step explanation:

It has something to do with the Weierstrass substitution, where we have

$\int\, f(\sin(x), \cos(x))dx = \int\, \dfrac{2}{1+t^2}f\left(\dfrac{2t}{1+t^2}, \dfrac{1-t^2}{1+t^2} \right)dt$

First, consider the double angle formula for tangent:

$\tan(2x)= \dfrac{2\tan(x)}{1-\tan^2(x)}$

Therefore,

$\tan\left(2 \cdot\dfrac{x}{2}\right)= \dfrac{2\tan(x/2)}{1-\tan^2(x/2)} = \tan(x)=\dfrac{2t}{1-t^2}$

Once the double angle identity for sine is

$\sin(2x)= \dfrac{2\tan(x)}{1+\tan^2(x)}$

we know $\sin(x)=\dfrac{2t}{1+t^2}$ , but sure, we can derive this formula considering the double angle identity

$\sin(x)= 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)$

Recall

$\sin \arctan t = \dfrac{t}{\sqrt{1 + t^2}} \text{ and } \cos \arctan t = \dfrac{1}{\sqrt{1 + t^2}}$

Thus,

$\sin(x)= 2 \left(\dfrac{t}{\sqrt{1 + t^2}}\right) \left(\dfrac{1}{\sqrt{1 + t^2}}\right) = \dfrac{2t}{1 + t^2}$

Similarly for cosine, consider the double angle identity

Thus,

$\cos(x)= \left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2- \left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2 = \dfrac{1}{t^2+1}-\dfrac{t^2}{t^2+1} =\dfrac{1-t^2}{1+t^2}$