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3 years ago

7

Underline each term in the following polynomials and write the number, variable or expression that each term has in common in th

e space provided

1 answer:

olga2289 [7]3 years ago

4 0

What are the polynomials

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Draw a number line -5 through 5 and mark 2x<6

inna [77]

Answer:

$x < 3$

Step-by-step explanation:

Before number line can be drawn, there is need to find the value of x in the inequality given. Thus,

$2x < 6$

Divide both sides by 2

$\frac{2x}{2} < \frac{6}{2}$

$x < 3$

The number line is indicated in the attached picture.

8 0

4 years ago

Please help click on the image below

Shtirlitz [24]

I believe that is c if I'm wrong I'm very sorry

6 0

3 years ago

To find the extreme values of a function f(x,y) on a curve xequalsx(t), yequalsy(t), treat f as a function of the single vari

pychu [463]

Answer:

Absolute maximum is 2

Absolute minimum at -2

Step-by-step explanation:

The given parametric functions are:

$x=2\cos t,y=2\sin t$

By the chain rule:

$f'(t)=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$\frac{df}{dt} =\frac{2 \cos t}{-2\sin t} =-\cot(t)$

At fixed points, $f'(t)=0$

$\implies -\cot (t)=0$

This gives $t=\frac{\pi}{2} ,\frac{3\pi}{2}$ on $0\le t\le 2\pi$

This implies that the extreme points are $(2\cos \frac{\pi}{2}, 2\sin \frac{\pi}{2})=(0,2)$ and $(2\cos \frac{3\pi}{2}, 2\sin \frac{3\pi}{2})=(0,-2)$

By eliminating the parameter, we have $x^2+y^2=4$

This is a circle with radius 2, centered at the origin.

Hence (0,2) is an absolute maximum ,at $t=\frac{\pi}{2}$ and (0,-2) is an absolute minimum at $t=\frac{3\pi}{2}$

7 0

3 years ago

HELP ASAP PLEASE.

Schach [20]

Answer:

5(6)81 because the are

Step-by-step explanation:

6•82

8 0

3 years ago

Which of the following numbers belong to the set of complex numbers AND The set of imaginary numbers? Select all that apply.

Schach [20]

<span>To be both complex and imaginary, the answer would have to include "i" as well as have the form "a + bi". The first example, 3i - 5, fits. This would look to be the only example that does fit. The negative square root of 14 would be imaginary, but not complex.</span>

7 0

3 years ago