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Tamiku [17]
2 years ago
5

If it is a regular polygon is then it is equilateral

Mathematics
2 answers:
Readme [11.4K]2 years ago
7 0
That's true. If it is a regular polygon, then all of its sides have the same length.
OLEGan [10]2 years ago
4 0
This might be a t or f. if this would be true,
regular polygons all contain an equal length in each side, polygons in general sometimes don't have equal sides. however, another kind of polygons are regular ones. those are polygons that use specific number of sides with the length of each one being the same
if it is false, polygons in general dont have equal sides.
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What is the pay earned for 27 hours at $4.65 per hour?
Stella [2.4K]

Answer:

$125.55

Step-by-step explanation:

All you do is multiply your earned per hour by total hours.

6 0
2 years ago
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In a cafeteria,
____ [38]

Answer:

2 students

Step-by-step explanation:

First, you find the number of students eating either salads and sandwiches. Then, you add the number of students eating salads and sandwiches together. Finally, you will subtract that number from the 12 total students

<h3>2/3 * 12/1 = 8 (sandwiches)</h3><h3>1/6 * 12/1 = 2 (salads)</h3><h3>8 + 2 = 10 (combined)</h3><h3>12 - 10 = 2 students</h3><h3 />
7 0
2 years ago
There are 10 multiple choice questions in an examination. first 4 have 5 choices each and remaining have 4 choices each. how man
stira [4]

2560000 sequences of answers are possible.

<em><u>Explanation</u></em>

Total number of questions = 10

First 4 questions have 5 choices each and remaining (10-4)= 6 questions have 4 choices each.

So, the possible sequences of answers for first 4 questions = 5^4 = 625

and the possible sequences of answers for remaining 6 questions = 4^6=4096

Thus, the total possible number of sequences of answers = (625 × 4096) = 2560000

5 0
3 years ago
Read 2 more answers
Help plz I NEED IT Help help hjelp
DaniilM [7]

Answer:

12

Step-by-step explanation:

12 because that's how many dots/vertices are shown(sorry if im wrong)

6 0
2 years ago
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If lim x-&gt; infinity ((x^2)/(x+1)-ax-b)=0 find the value of a and b
MAXImum [283]

We have

\dfrac{x^2}{x+1}=\dfrac{(x+1)^2-2(x+1)+1}{x+1}=(x+1)-2+\dfrac1{x+1}=x-1+\dfrac1{x+1}

So

\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-ax-b\right)=\lim_{x\to\infty}\left(x-1+\frac1{x+1}-ax-b\right)=0

The rational term vanishes as <em>x</em> gets arbitrarily large, so we can ignore that term, leaving us with

\displaystyle\lim_{x\to\infty}\left((1-a)x-(1+b)\right)=0

and this happens if <em>a</em> = 1 and <em>b</em> = -1.

To confirm, we have

\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-x+1\right)=\lim_{x\to\infty}\frac{x^2-(x-1)(x+1)}{x+1}=\lim_{x\to\infty}\frac1{x+1}=0

as required.

3 0
2 years ago
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