All Answers 358468 Answers

1 = 11
2 = 142
3 = 273
4 = 404
5 = 535
6 = 666
7 = 797
8 = 928
9 = 1059
10 = 1190
11 = 1321
12 = 1452
13 = 1583
14 = 1714
15 = 1845
16 = 1976
17 = 2107
18 = 2238
19 = 2369
20 = 2500
21 = 2631
22 = 2762
23 = 2893
24 = 3024
25 = 3155
26 = 3286
27 = 3417
28 = 3548
29 = 3679
30 = 3810
31 = 3941
32 = 4072
33 = 4203
34 = 4334
35 = 4465
36 = 4596
37 = 4727
38 = 4858
39 = 4989
 

Compute a^3 + b^3.

Hello Guest!

\(2x^2 - 8x + 7 = x^2 - 6x + 1\\ x^2-2x+6=0\\ x=1\pm \sqrt{1-6}\\ \color{blue}x\in \{1-i\sqrt{5},1+i\sqrt{5}\}\\ a=1-i\sqrt{5}\\ b=1+i\sqrt{5}\\a^3=-14+2\sqrt{5}i\\ b^3=-14-2\sqrt{5}i\\ \color{blue}a^3+b^3=-28\)

  !

DX = 11

  !

b + c = -12.

The solution is (2,-3,5,-1,8).

The sum is 5498.

1 is 2044

2 is 19

3 is 541

wrong

wrong

The sum of the first 20 terms is 308.

1) The sum is 4055

2) There are 44 terms.

3) c = 182.

thats wrong

A + B = -13/2.

1 is 2013

2 is 17

3 is 501

The answer is 10x - 4y = 32

The remainder is 61.

From the formula for R, the circumradius is 150/7.

Anna had 324 dollars at first.

This is obvious if you use geometric series.

Cube the monomial (2b6m4)3

(2b^6m^4)^3

\((2b^6m^4)^3\\ =2^3(b^6)^3(m^4)^3\\ =8b^{18}m^{12}\)

Thanks Alan,

That was good revision for me. :)

See https://web2.0calc.com/questions/pre-calc-problem#r5

Thanks guest.

14 is correct.  I think guest did it with a calculator.

This is fine but I will give a non-calculator answer

Find N such that  27*N (mod29) = 1

\(27\equiv-2 \mod 29\\ 30\equiv 1 \mod 29\\ -2*-15 = 30 \equiv 1 \mod29\\ -15\equiv 29-15 \mod29 \equiv 14 \mod29\\ so\\ 27*14=1 \mod29\\ \text{So 14 is the modular inverse of 27 (mod 29)}\)

You question makes no sense to me.  

For instance what is   b.0.  ??