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PEDMAS and BODMAS, it depends on what country you are in.

ROUND THE ANSWER TO 

TWO DECIMAL POINTS!

I don't really understand the question.

3-x3x+2

What are the x's supposed to be here?

If they are multiplication signs, I don't understand what the minus sign in front of it means.

The sum, Sn,  of the first n terms of a geometric series = 

Sn  = a1  ( 1 - r^n) / ( 1  - r)   where a1 is the first term and r is the common ratio

484  = a1  ( 1 - 3^n)  / (1 -3)

484  =  a1 ( 1 - 3^n) / -2

-968  = a1 ( 1 - 3^n)      rearrange  as

a1  =   -968 / ( 1 - 3^n)    factor out a negative

a1  =  968 /(3^n - 1)         (1)

And  the nth term is given by

an  = a1 (3)^(n -1)

324  = a1 * (3)^(n - 1)       sub (1)  into this for  a1

324  = 968 / (3^n - 1) * 3^(n - 1)         .......3^(n - 1)   =  3^n / 3

324 (3^n - 1)  =  968  (3^n) / 3

972 (3^n - 1)  = 968 (3^n)

972 (3^n)  - 972   = 968 (3^n)

972(3^n) - 968(3^n)  =  972

4(3^n)  = 972

3^n  =  243        ..........243  = 3^5

3^n  = 3^5

So.....n = 5

So....using (1)

The first term, a1  = 

968 / (3^n - 1)

968/ ( 3^5 - 1)

968 / (243 - 1)

968 / 242 =

4

Let a1  be the first term, am be the mth term, an be the nth term and d be the common difference between terms

The sum of the first m terms =

m^2 =  m/2 * (a1  + am)  → 2m  =  a1 + am   (1)

The sum of the first n terms  =      

n^2  =  n/2 * (a1 + an)  →  2n  =  a1  + an    (2)

Subtract  (2)  from (1)

2m - 2n = am - an     (3)  

And the  mth term is given by 

am =  a1 + d (m - 1)  →  am  = a1 + dm - d   (4)

And the nth term is given by

an = a1 + d(n - 1) →  an  = a1 + dn - d     (5)

Subtract  (5) from (4)

am - an   =  dm - dn     (6)

Set (3)  and (6)   equal to solve for d

2m - 2n   =  dm - dn

2(m - n)   =  d (m - n)

2 = d

Rearrange (1)   and solve for a1

am   = a1 + 2m - 2

2m - a1 = a1 + 2m - 2

2m = 2a1 + 2m - 2

0 = 2a1 - 2

2 = 2a1         divide by 2

1  = a1

Using (4)  and (5)  and subbing for a1  and  d

am  =  1 + 2m - 2 →  am  = 2m - 1

an =   1   + 2n - 2  →  an  = 2n - 1

So.....the ratio of the mth term  to the nth term   =

am : an  =    2m - 1   :  2n - 1 

We can use the tangent inverse......

arctan ( 25 / 45)  = D  ≈   29.05°

We can use the cosine...

cos (55°)  =  34 / L        where L is the length of the guy wire

Rearrange as

L  = 34 / cos (55°)  ≈  59.2 ft   = 59 ft

Start by combining the negative and the positive numbers. Then solve.

2(x-1)=14

14/2=7

7+1=8

x=8

no offence but i'd like it if one of the moderators answered me ok?

Are you 100% sure its solveable?

i appreciate people that come up with their own questions, but it might be a very hard question, a question im afraid im not able to solve.

 p^2+q^2 +r^2

sorry they are not multiple

a prime is a number that is not divisble by any number.

2*r+2*q+2*p can sometimes be a prime, according to you.

are you sure 2*r+2*q+2*p can be a prime? think about it (its a hint)

2(x-1)=14

(2*x)+(2*-1)=14

2x-2=14

+2       +2

2x=16

/2     /2

x=8

beta=B

sec(B)=1/(cos)

tan(B)=sin(B)/cos(B)

tan^2(B)=5-4sec(B) ......multiply by cos^2(B)

sin^2(B)=5*cos^2(B)-4cos(B)

using the identity sin^2(X)+cos^2(X)=1

sin^2(B)=1-cos^2(B)=5cos^2(B)-4*cos^2(B)........add (cos^2(B)-1)

0=6*cos^2(B)-4cos^2(B)-1.......divide by 6

0=cos^2(B)-(2/3)*cos(B)-1/6=(cos(B)-(1/3))^2-1/18......add 1/18

1/18=(cos(B)-(1/3))^2

i think you can continue from here

are c, e and p constants? if not then there is no answer my friend

dipende da cosa polinomio si sta parlando. Io non credo che ci sia un modo per fare questo per ogni polinomiale. è possibile utilizzare la formula quadratica di farlo con un'equazione di secondo grado. Spero che aiutato!

Unknown.

I cant understand your question, please make it more clear

what is your question then?

Three statisticians went out hunting, and came across a large deer. The first statistician fired, but missed, by a meter to the left. The second statistician fired, but also missed, by a meter to the right. The third statistician didn't fire, but shouted in triumph, "we got it!"
 

A statistics major was completely hung over the day of his final exam. It was a True/False test, so he decided to flip a coin for the answers. The stats professor watched the student the entire two hours as he was flipping the coin...writing the answer...flipping the coin...writing the answer. At the end of the two hours, everyone else had left the final except for the one student. The professor walks up to his desk and interrupts the student, saying:
"Listen, I have seen that you did not study for this statistics test, you didn't even open the exam. If you are just flipping a coin for your answer, what is taking you so long?"

The student replies bitterly, as he is still flipping the coin: "Shhh! I am checking my answers!"

1 817829 852777 597057 552405 801557 063757 035128 314603 081443 374215 399132 611377 574662 310544 622023 540398 413816 236264 614053 001202 262619 126842 307342 323681 (145 digits) = 7 × 151 × 211 × 17351 × 196439 × 439171 × 33434 876129 × 1 976924 700193 × 82379 672980 635330 027596 216115 831844 876154 751458 947502 203898 014052 739376 120879 214393 428435 681495 644521 (101 digits).

P.S. The last factor of 101 digits is also a composite number, but it may take a long time to factor it, for it appears to have 2 large factors of about 50-digits each!.

Atleast there is a reliable source in this site then :D

You dont even seem to try and you are pretty much spamming the questions list but ill help you:

S_m=(a+(a+d*(m-1)))*m/2

S_n=(a+(a+d*(n-1)))*n/2

S_m/S_n=(m/n)*(2a+d(m-1))/(2a+d(n-1))=m²/n²

Therefore, (2a+d(m-1))/(2a+d(n-1))=m/n

What you need to do is to show m=n or 2a=d, then ill show you the answer.

Atleast try to do something

I DID'NT ask any questions of the form prove that if. and i do know that prove that if questions has only 1 answer.

Sir, that doesn't MATTer.