The solution gives both A and M. I suggest you read over the solution, heureka did an amazing job on it :)
6+b+c = 15
b+c = 15
b = 15 - c
6^2 + b^2 = c^2
plug in b:
6^2 + (15-c)^2 = c^2
36+225+c^2-30c = c^2
cancel out the c^2 and combine constants
261 - 30c = 0
30c = 261
c = 261/30
:)
Already been answered I believe: https://web2.0calc.com/questions/the-solution-of-8x-1-5-mod-12-is-x-a-mod-m-for-some
I'm back babyyyyyy
The equation for the sum of the infinite geometric series is S = a1/(1-r).
We know S and the ratio, so we plug in: 25 = a1/(1-(-1/3)), 25 = a1/4/3, 25 = 3a1/4, 100 = 3a1, a1 = 100/3.
The next term of the sequence is 100/3 times -1/3, which is -100/9
Find the surface area of both
I calculate the difference to be 72 in^2
Multiply by 30: 72*30 = 2160
x = 1, 2, -2isqrt(2), 2isqrt(2)
Acellular slime molds do not have cell walls. As a result, the nuclei are combined into one giant clump of nuclei in the cytoplasm.
13/20 is equal to 78/120, so 78 people have dogs.
Let A be the number of adult tickets and C be the amount of children tickets. We can form the equation 105A + 60C = 7155, as we converted it from dollars to regular integers. We can simplify the equation by dividing by 5, giving 21A + 12C = 1431. Dividing by 3 gives 7A + 4C = 477. 477 divided by 7 gives 68 remainders 1, and if we remove one more seven and add it to one, we will have a multiple of 4. This means 67 adult tickets and 2 children tickets.
C'mon man. First, at least simplify the latex. Also, it's literally rounding. If you do not know how to find the answer to a problem like this, then you need to learn how to, not just ask to get an answer to your AoPS thing or whatever.
