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What is the product of all the divisors of 1000?

Ignoring the number 1:  1000  =  2 x 2 x 2 x 5 x 5 x 5  =  2³ x 5³

The factors will be:  2,   2²,   2³,   5,   5²,   5³,   2·5,   2·5²,   2·5³,   2²·5,   2²·5²,   2²·5³,   2³·5,   2³·5²,   2³·5³   

Multiplying all of these together:  2²⁴ x 5²⁴  

999  -  100 +  1 = 900 numbers between 100 and 999 inclusive.

There are 252 numbers with at least one 7 in them between 100 and 999 inclusive.

There are: 900  - 252 = 648 numbers that have NO 7 in them between 100 and 999 inclusive.

I commented on your other post, it is important.

uh......................... this is science, not math, this is a math forum, if you need to any help with your math, I am happy to help you the best I can:)

2197/20825

any ideas?

Darn it, I missed a sign when I started the answer. 

It should be:  

y = y so you can set the equivalents equal 

                                                       x² – 2  =  –x + 4 

subtract  –x + 4 from both sides      x² + x – 6  =  0   

factor the quadratic                         (x – 2)(x + 3)  =  0 

                                                        x = +2  and  x = –3 

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Who gave me that negative heart?  I haven't seen YOU contributing to the solution.  I understand now how I didn't get it right the first time - an oversight, obvious in hindsight - but I stuck with it and kept coming back and I think I finally got the correct answer.   

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Understand that there is one job. 

Fred works at a rate of   1  job                   2  job 

                                      ———   call this   ———   to enable you to add it to John's rate

                                        3  hr                    6  hr 

John works at a rate of   1  job 

                                      ——— 

                                        6  hr 

Add them together and you get   3  job                  1 job 

                                                   ———  which is   ———    together they do the job in 2 hours   

                                                     6 hr                     2 hr 

This kind of problem was so hard for me when I was in school.  It took me forever to grasp how to do it. 

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what is y=x2−2 and y=−x+4 solutions  

y = y so you can set the equivalents equal 

                                                    x² – 2 = x + 4 

subtract x + 4 from both sides      x² – x – 6 = 0   

factor the quadratic                      (x + 2)(x – 3) = 0 

                                                     x = –2  and  x = 3 

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The answer has to be 6. 

See  https://web2.0calc.com/questions/help-me-asap-please_8  for my reasoning. 

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If you take one exterior angle at each vertex of any polygon (triangle, quadrilateral, ...), their sum will always 360^o.

So, the sum of 158^o, 99^o, and the unknown angle is 360^o.

--->   158 + 99 + x  =  360  ... solve for x ...

"I already tried that. it isn't right."

It's me again.  I had an idea ... and it's so crazy it just might work. 

It occurred to me that when you have a negative number to start with,

raising it to an even-numbered exponent produces a positive result. 

So here's my idea:  Try 6 and see if that gets accepted. 

If it works, it's because these are the other two:  (–8)^–2  and  (–2)^–6  

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      Compute (1 + sqrt(3)*i)^10.                   

243 i^10 + 810 sqrt(3) i^9 + 3645 i^8 + 3240 sqrt(3) i^7 + 5670 i^6 + 2268 sqrt(3) i^5 + 1890 i^4 + 360 sqrt(3) i^3 + 135 i^2 + 10 sqrt(3) i + 1    (11 terms)

1/3  +  1/6 =1/J

J = 2 hours working together.

Nature photojournalist James Balog founded the Extreme Ice Survey in 2007 after spending much of the previous two years photographing receding glaciers for National Geographic and The New Yorker. Balog saw extraordinary amounts of ice vanishing with shocking speed.

Isn't this just E = Mc^2

0.01  -  0.009 =0.001 kg - loss in Mass

E = 0.001  x  299,792,458^2 = 89,875,517,873,681.764 =8.99  x  10^13 Joules

Note: I don't think you need Planck's constant in this calculation. It is generally used to calculate the energy of a photon of radiation when its frequency is known.