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grin007 [14]
3 years ago
7

Can i get a list of prime numbers

Mathematics
1 answer:
bija089 [108]3 years ago
5 0

well prime numbers are noting but whole number greater than 1 whose only factors are 1 and itself.

soo....the first few prime numbers are :

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149 etc.

and if u c .... all the numbers dont come in any other table rather that one and itself


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the cost to join a yoga studio is $100, plus $25 per month. What is the cost of the first 8 months at the yoga studio ?
umka21 [38]
25*8= 200
so $200 for 8 months
if you want the total add 200+100=300
4 0
3 years ago
Pls help with this asap!!!
katrin2010 [14]

4x4x4=64

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Answer = -44

3 0
3 years ago
Read 2 more answers
If a, b, c are all non -zero and a+b+c= 0, prove a²÷ bc +b²÷ ac +c²÷ ab =3
Step2247 [10]

Combining the fractions gives

\dfrac{a^2}{bc}+\dfrac{b^2}{ac}+\dfrac{c^2}{ab}=\dfrac{a^3+b^3+c^3}{abc}

Since a+b+c=0, we also have

(a+b+c)^3=0

a^3+b^3+c^3+3a^2b+3a^2c+3ab^2+3b^2c+3ac^2+3bc^2+6abc=0

If we add and subtract the last 7 terms of the left hand side to/from the numerator, we get

\dfrac{(a+b+c)^3-(3a^2b+3a^2c+3ab^2+3b^2c+3ac^2+3bc^2+6abc)}{abc}

=-3\dfrac{a^2b+a^2c+ab^2+b^2c+ac^2+bc^2+2abc}{abc}

=-3\dfrac{(a+b)(a+c)(b+c)}{abc}

Also because a+b+c=0, we have

\begin{cases}a=-(b+c)\\b=-(a+c)\\c=-(a+b)\end{cases}

so we find that

-3\dfrac{(a+b)(a+c)(b+c)}{abc}=-3\dfrac{(-c)(-b)(-a)}{abc}=3\dfrac{abc}{abc}=3

4 0
4 years ago
Help asappppp<br> Quickkkkkkkkkkk
kicyunya [14]

Answer:

The answer is C

Step-by-step explanation:

In these questions you have to remember crawl before you walk. So x First then y match up the coordinates (-4,9) and such and the answer should reveal itself.

Regards,

Fellow 8th grader, Abdelmumen Nakoa

8 0
3 years ago
Maria uses 1 1/2 feet of string to make one bracelet . how much string will she use to make 5 bracelets?
Neko [114]

Answer:

7 1/2 feet of string to make 5 bracelets

can you also please mark me as brainliest?

8 0
3 years ago
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