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3 years ago

How to find the degree in the polynomial

Mathematics

1 answer:

Tatiana [17]3 years ago

6 0

Answer:

To find the degree of the polynomial you need to add up the exponents of each term and pick the highest sum.

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A right rectangular pyramid is sliced vertically (down) by a plane passing

Anton [14]

A right rectangular pyramid when sliced vertically, the shape of the cross-section is known as Triangle.

<h3>What is A triangle?</h3>

This is known to be a kind of shape that is said to be in a closed form and it is also known to be a 2-dimensional shape that has 3 sides, 3 angles, and also 3 vertices.

Note that when the when the right rectangular pyramid is sliced vertically (down) by a plane passing through the of the pyramid, the new shape of the cross-section is a triangle.

See full question below

A right rectangular pyramid is sliced vertically (down) by a plane passing through the of the pyramid. What is the shape of the cross-section?

A. Rectangle

B. Pyramid

C. Triangle

D. Trapezoid

See full question below

Learn more about triangle from

brainly.com/question/17335144

#SPJ1

3 0

2 years ago

Joaquin is constructing the perpendicular bisector of ab. What should be his first step?

natali 33 [55]

Answer:

D. Open the compass so that the distance from the two points of the compass is wider than half the length of $\over{AB}$ .

Step-by-step explanation:

To construct a perpendicular for $\over{AB}$ , we must first take a compass & take the distance of its arms wider than half the length of $\over{AB}$ .

This is done in order to get two intersecting arcs in the top & bottom of $\over{AB}$ so that a perpendicular bisector can be drawn through it.

After two intersecting lines are drawn below & above $\over{AB}$ , draw a line joining these 2 points through their points of intersection. The point where it intersects $\over{AB}$ is the middle-most point of $\over{AB}$ & now a perpendicular bisector of $\over{AB}$ is constructed.

$\rule{150pt}{2pt}$

3 0

2 years ago

You measured 6 cups of water and 5 cups of lemon juice to make how many total quarts of lemonade? quarts cups

crimeas [40]

1 quart is 4 cups so 11 cups is 2.75 quarts

7 0

3 years ago

Read 2 more answers

What is the domain and the range of these functions?? quick please!!!!!!!!!!

Sever21 [200]

The domain of a function f(x) is the set of all values for which the function is defined, and the range of the function is the set of all values that f takes. A rational function is a function of the form f(x)=p(x)q(x) , where p(x) and q(x) are polynomials and q(x)≠0 .

6 0

3 years ago

For <img src="https://tex.z-dn.net/?f=e%5E%7B-x%5E2%2F2%7D" id="TexFormula1" title="e^{-x^2/2}" alt="e^{-x^2/2}" align="absmiddl

nevsk [136]

I'm assuming you're talking about the indefinite integral

$\displaystyle\int e^{-x^2/2}\,\mathrm dx$

and that your question is whether the substitution $u=\dfrac x{\sqrt2}$ would work. Well, let's check it out:

$u=\dfrac x{\sqrt2}\implies\mathrm du=\dfrac{\mathrm dx}{\sqrt2}$
$\implies\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt2\int e^{-(\sqrt2\,u)^2/2}\,\mathrm du$
$=\displaystyle\sqrt2\int e^{-u^2}\,\mathrm du$

which essentially brings us to back to where we started. (The substitution only served to remove the scale factor in the exponent.)

What if we tried $u=\sqrt t$ next? Then $\mathrm du=\dfrac{\mathrm dt}{2\sqrt t}$ , giving

$=\displaystyle\frac1{\sqrt2}\int \frac{e^{-(\sqrt t)^2}}{\sqrt t}\,\mathrm dt=\frac1{\sqrt2}\int\frac{e^{-t}}{\sqrt t}\,\mathrm dt$

Next you may be tempted to try to integrate this by parts, but that will get you nowhere.

So how to deal with this integral? The answer lies in what's called the "error function" defined as

$\mathrm{erf}(x)=\displaystyle\frac2{\sqrt\pi}\int_0^xe^{-t^2}\,\mathrm dt$

By the fundamental theorem of calculus, taking the derivative of both sides yields

$\dfrac{\mathrm d}{\mathrm dx}\mathrm{erf}(x)=\dfrac2{\sqrt\pi}e^{-x^2}$

and so the antiderivative would be

$\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt{\frac\pi2}\mathrm{erf}\left(\frac x{\sqrt2}\right)$

The takeaway here is that a new function (i.e. not some combination of simpler functions like regular exponential, logarithmic, periodic, or polynomial functions) is needed to capture the antiderivative.

3 0

3 years ago