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3 years ago

How does the shape of a cross-section of a 3D figure parallel to the base relate to the volume of the 3D figure?

Mathematics

1 answer:

Ket [755]3 years ago

3 0

Answer:

The shape of each cross-section of a 3D figure, relates to the volume because the area of the cross-section is determined by its shape and the area of this cross section is in the sum that calculates the volume of this 3D figure.

Step-by-step explanation:

An infinite sum of all the all the cross-sections of a 3D figure parallel to the base equals the volume of that 3D figure.

You might be interested in

Find the sum of the given vectors and illustrate geometrically. [-1,4],[6,-2]

cricket20 [7]

Answer:

The sum of vectors is <5,2>.

Step-by-step explanation:

The given vectors are <-1,4> and <6,-2>.

We need to find the sum of the given vectors and illustrate geometrically.

Plot the point (-1,4) on a coordinate plane and draw a vector <a> from (0,0) to (-1,4).

Plot the point (6,-2) on a coordinate plane and draw a vector <b> from (0,0) to (6,-2).

Now complete the parallelogram and the diagonal represents the sum of both vectors.

The end point of the diagonal is (5,2). It means sum of vectors is <5,2>.

Check the sum of vectors:

$+==$

Therefore, the sum of vectors is <5,2>.

4 0

4 years ago

Help!!!!PLEASE & THANKS

kiruha [24]

x° = 25°

25° is the answer

7 0

2 years ago

<img src="https://tex.z-dn.net/?f=%28x%5E%7B2%7D%20%2Bx-3%29%3A%20%28x%5E%7B2%7D%20-4%29%5Cgeq%201" id="TexFormula1" title="(x^{

Jlenok [28]

Answer:

$x>2$

Step-by-step explanation:

When given the following inequality;

$(x^2+x-3):(x^2-4)\geq1$

Rewrite in a fractional form so that it is easier to work with. Remember, a ratio is another way of expressing a fraction where the first term is the numerator (value over the fraction) and the second is the denominator(value under the fraction);

$\frac{x^2+x-3}{x^2-4}\geq1$

Now bring all of the terms to one side so that the other side is just a zero, use the idea of inverse operations to achieve this:

$\frac{x^2+x-3}{x^2-4}-1\geq0$

Convert the (1) to have the like denominator as the other term on the left side. Keep in mind, any term over itself is equal to (1);

$\frac{x^2+x-3}{x^2-4}-\frac{x^2-4}{x^2-4}\geq0$

Perform the operation on the other side distribute the negative sign and combine like terms;

$\frac{(x^2+x-3)-(x^2-4)}{x^2-4}\geq0\\\\\frac{x^2+x-3-x^2+4}{x^2-4}\geq0\\\\\frac{x+1}{x^2-4}\geq0$

Factor the equation so that one can find the intervales where the inequality is true;

$\frac{x+1}{(x-2)(x+2)}\geq0$

Solve to find the intervales when the equation is true. These intervales are the spaces between the zeros. The zeros of the inequality can be found using the zero product property (which states that any number times zero equals zero), these zeros are as follows;

$-1, 2, -2$

Therefore the intervales are the following, remember, the denominator cannot be zero, therefore some zeros are not included in the domain

$x\leq-2\\-2$

Substitute a value in these intervales to find out if the inequality is positive or negative, if it is positive then the interval is a solution, if it is negative then it is not a solution. This is because the inequality is greater than or equal to zero;

$x\leq-2$ -> negative