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∑[2 + 4 + 6 + ... + 100 + 102 + 104]==2,756 mod 7==5 - the remainder.

2+4+6...+100+102+104 is equal to 10920. The remainder of 10920 divided by 7 is 0. 

So the answer is 0

Let austin's age be a.

3 * (a - 3) = a + 18

3a - 9 = a + 18

continue solving after that :)

It goes upstream at 10km/hour for 3 km so that takes  0.3 of an hour

It goes downstream tat  18km/hour for 3km so that takes  1/6 of an hour

Total distance = 6km    total time = (18+10)minutes = 28/60 = 0.4667 hour

ave speed = 6/0.4667 = 12.86 km/hour

The first letter has 26 possibilities. The second has 5, and the third has 21. 26*5*21 = 2730, so the answer is 2730

14 - 4 ==10 km/h - its speed upstream

3 km / 10 km/h ==3/10 x 60 ==18 minutes going upstream

14 + 4 ==18 km/h - its speed downstream

3 km/18 km/h ==1/6 x 60 ==10 minutes going downstream

Average speed ==total distance travelled / time

                            == [3 * 2] / [18 + 10] / 60

                            ==  6 / [28 / 60]

                            == 6  * 60 / 28 ==360 / 28 ==12 6/7 km/h

It goes upstream at 10km/hour for 3 km so that takes  0.3 of an hour

It goes downstream tat  18km/hour for 3km so that takes  1/6 of an hour

Total distance = 6km    total time = (18+10)minutes = 28/60 hour

ave speed =   \(6\div \frac{28}{60}=12\frac{6}{7}\;km/hour\)

Sorry guest answerers, I was correcting my answerer without seeing that you had already done so.

I did make a very careless error and I thank you for correcting it.

Hint:    Start by solving this to find n.

\( (x^{0.5})^n*(\frac{1}{x^2})^{10-n}=x^0\\\)

there is nothing to solve.

This cannot be the whole question or else it simply is not a question.

A + B + C ==1512 necklaces.

A / 1512 ==2 / 7, solve for A

A ==432 Necklacesin box A

1512   -  432 ==1,080 Necklaces in box B + C

[B - 190] ==Necklaces left in box B

[C  - 1/4C] ==3/4C - Necklaces left in box C

B  +  C  ==1080..........................(1)

[B - 190] / [3/4C] ==2/1............(2), solve for B, C

B==724  and  C ==356 - number of necklaces in box C at first.

You answer has no credibility whatsoever. So please justify it or go away.

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Use the quadratic formula above to solve for x:

x ==2

See https://web2.0calc.com/questions/graphing-question_40

The minimum value of x^2 + y^2 is 16.

17^(77^2) mod 35 ==17 - the remainder

2013^(2013^2) mod 13 ==8 - the remainder.

(1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 27, 30, 32, 36, 40, 45, 48, 54, 60, 64, 72, 80, 81, 90, 96, 108, 120, 135, 144, 160, 162, 180, 192, 216, 240, 243, 270, 288, 320, 324, 360, 405, 432, 480, 486, 540, 576, 648, 720, 810, 864, 960, 972, 1080, 1215, 1296, 1440, 1620, 1728, 1944, 2160, 2430, 2592, 2880, 3240, 3888, 4320, 4860, 5184, 6480, 7776, 8640, 9720, 12960, 15552, 19440, 25920, 38880, 77760)>>Total = 84 divisors.

(2, 4, 6, 8, 10, 12, 16, 18, 20, 24, 30, 32, 36, 40, 48, 54, 60, 64, 72, 80, 90, 96, 108, 120, 144, 160, 162, 180, 192, 216, 240, 270, 288, 320, 324, 360, 432, 480, 486, 540, 576, 648, 720, 810, 864, 960, 972, 1080, 1296, 1440, 1620, 1728, 1944, 2160, 2430, 2592, 2880, 3240, 3888, 4320, 4860, 5184, 6480, 7776, 8640, 9720, 12960, 15552, 19440, 25920, 38880, 77760) >>Total = 72 EVEN DIVISORS.

13  -  4  == 9 days' work left

3/5 of the work force left

9 / ( 3/5) ==9  x  5/3 ==45 / 3 ==15 days to finish the software application.

OR: 

[5  x  13] / 3 ==21 2/3 days to finish the 13-day job.

21 2/3 x 9/13 [work left] ==15 days to finish the job.

Not funny

$DX = 5/2 \cdot \pi - 5$.

Hello I can't help you but I want to tell you that you can do it I believe in you have a good day

You can do it

AB ≈ 7.534                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                 

Yes i will post the answer very soon, thank you for answering it!