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Nikolay [14]
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
Mathematics
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 xequals​x(t), yequals​y(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
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