1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
d1i1m1o1n [39]
3 years ago
10

Why is

Mathematics
1 answer:
zloy xaker [14]3 years ago
3 0

Answer:

Only natural numbers (i.e., non-negative integers) can be the exponents of variables in a polynomial.

Step-by-step explanation:

The exponent of variables in a polynomial should be natural numbers (0, 1, 2, 3, \dots.)

  • \sqrt{2\, x} is equal to \sqrt{2}\, x^{1/2}. In this expression, x is the variable. Its exponent is 1/2, which isn't a natural number.
  • On the other hand, \sqrt{2}\, x is equivalent to \sqrt{2}\, x^{1}. The exponent of variable x is 1, which is indeed a natural number.

\sqrt{2\, x} isn't a polynomial because the exponent of variable x isn't a natural number. On the other hand, \sqrt{2}\, x is indeed a polynomial over the set of real numbers.

You might be interested in
What's the answer? I need help!
Juli2301 [7.4K]

Answer:

<h2><u>I can do algebra forever</u></h2>

First we can use distributive property. 2a-10 = 6. So we add 10 on both sides and get 2a=16. Then we divide both sides by 2 so we get a = 8. Therefore proved algebraically that a = 8. I can always do algebra. Feel free to keep asking questions like these. Well feel free to ask any question because this is brainly.

<h2><u>Answer is a = 8, 8 is answer</u></h2><h2><u></u></h2>

<u>brainliest</u>

8 0
3 years ago
Read 2 more answers
What the area of the triangle!<br><br><br> need help with these question would mean a lot
zloy xaker [14]
Heyy I don’t think your pictures are loading it just shows grey
3 0
3 years ago
Ccccc<br>ccccc<br>ccccc<br>ccccc<br>ccccc
makvit [3.9K]
Cccccccccccccccccccccccccccccccc
4 0
3 years ago
Coach Rivas can spend up to $750 on 30 swimsuits for the swim team. The inequality shown can be used to find the maximum amount
777dan777 [17]

Answer:

$0 < p ≤ $25

Step-by-step explanation:

We know that coach Rivas can spend up to $750 on 30 swimsuits.

This means that the maximum cost that the coach can afford to pay is $750, then if the cost for the 30 swimsuits is C, we have the inequality:

C ≤ $750

Now, if each swimsuit costs p, then 30 of them costs 30 times p, then the cost of the swimsuits is:

C = 30*p

Then we have the inequality:

30*p ≤ $750.

To find the possible values of p, we just need to isolate p in one side of the inequality.

So we can divide both sides by 30 to get:

(30*p)/30 ≤ $750/30

p ≤ $25

And we also should add the restriction:

$0 < p ≤ $25

Because a swimsuit can not cost 0 dollars or less than that.

Then the inequality that represents the possible values of p is:

$0 < p ≤ $25

6 0
3 years ago
I’m not sure how to do this problem
strojnjashka [21]
The first one would be x^12
6 0
3 years ago
Other questions:
  • I NEED HELP :_( Which of the following points is in the solution set of y &gt; -x∧2 + 5? (0, 5) (2, 4) (1, 3)
    5·1 answer
  • Find the coordinates of the image of point R(5, –4) rotated 90 degrees about the origin.
    15·1 answer
  • .If the product of 5 consecutive integers is not zero,which of following statements individually sufficient provide(s) whether t
    11·1 answer
  • Joe receives two dollars plus a dollar for each box he packs. Michael receives two dollars times the number of boxes he packs.
    15·2 answers
  • Simplify each ratio to lowest terms.
    15·2 answers
  • Which is the solution to the system of equations?
    11·1 answer
  • 7 1/8-3 2/3<br><br> As a mixed number in simplest form.
    8·2 answers
  • Arti bought 12 kilogram of sweets. How many<br> grams did she buy?
    5·1 answer
  • Please someone help me asap. Im giving brainliest !!!!
    11·2 answers
  • PLEASE HELP ME ASAPPPPPP!!!!!!!<br><br><br><br> TYSM!! :D
    15·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!