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allochka39001 [22]
3 years ago
5

Is 3/12 a rational number.

Mathematics
1 answer:
lara [203]3 years ago
6 0

yes it's rational because it equals .25 which terminates

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Which Inequality is represented by the number line graph?
USPshnik [31]

Answer:

option C is your right answer

Step-by-step explanation:

have a nice day!!

8 0
3 years ago
Sam invests $800 at 6%/a compounded annually for 5 years. What is the total interest<br> earned?
s2008m [1.1K]
<h3>Answer:    270.58 dollars</h3>

======================================================

Work Shown:

  • A = account value after t years
  • P = principal or amount deposited = 800
  • r = interest rate in decimal form = 0.06
  • n = number of times we compound per year = 1
  • t = number of years = 5

So,

A = P*(1+r/n)^(n*t)

A = 800*(1+0.06/1)^(1*5)

A = 1070.58046208

A = 1070.58

After five years, the account will have $1,070.58 in it.

The amount of interest earned is A-P = 1070.58 - 800 = 270.58 dollars.

4 0
2 years ago
-5/6+ -5/6 please help me
ehidna [41]
(-5/6)+(-5/6) have the same denominator (number at bottom of fraction), so all you have to do is add the numerator(top number). So -5 +-5 =-10 . Now you have -10/6, which simplifies to -5/3. 
Hope this helps :)
5 0
3 years ago
Read 2 more answers
Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$
Viktor [21]

When n=0, we have

5^{3^0} + 1 = 5^1 + 1 = 6

3^{0 + 1} = 3^1 = 3

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for n=k, that

3^{k + 1} \mid 5^{3^k} + 1

Now for n=k+1, we have

5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)

so we know the left side is at least divisible by 3^{k+1} by our assumption.

It remains to show that

3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1

which is easily done with Fermat's little theorem. It says

a^p \equiv a \pmod p

where p is prime and a is any integer. Then for any positive integer x,

5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3

Furthermore,

5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3

which goes all the way down to

5^{3^k} \equiv 5 \pmod 3

So, we find that

\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3

QED

5 0
1 year ago
What is the lowest common denominator of 6,4,3,42?
TEA [102]

Answer: 1

Step-by-step explanation: the lowest number they divide into evenly is 1

8 0
2 years ago
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