catmg

Yayy :DDD

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Try to check your spam, and make sure your email is correct. :))

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I think you just have to solve for no zeros. 

No zeros: 9^50

Total: 9*10^49

9*10^49-9^50 

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Happy birthday Melody!!!! :DDDDD

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you're welcome :))

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 a = (4+9+8)^(-1)

a = 10^(-1)

a = 10

Hope this helped. :))

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15/21 = 5/7 = 10/14 = 15/21 = 20/28 = 25/35

Which one is wrong?

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Ohhh, I've never thought of using a triangle to solve these problems. 

It seems a lot easier to do it like that. 

Thank you. :))

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I am very confused too, but I'll try my best. 

question 1

cos(x) = sqrt(1-sine(x)^2)

cos(x) = sqrt(1 - (-6k/(k^2+9))^2)

cos(x) = sqrt(1-(36k^2)/(k^4+18k^2+81))

cos(x) = sqrt((k^4-18k^2+81)/(k^4+18^2+81))

cos(x) = (k^2-9)/(k^2+9)

question 2

tan(x)+1/cos(x) = tan(x)+1/cos(x)

tan(x)+1/cos(x) = sine(x)/cos(x) + 1/(cos(x))

tan(x)+1/cos(x) = (sine(x)+1)/cos(x)

tan(x)+1/cos(x) = (-6k/(k^2+9)+1)/((k^2-9)/(k^2+9))

tan(x)+1/cos(x) = ((k^2-6k+9)/(k^2+9))/((k^2-9)/(k^2+9))

tan(x)+1/cos(x) = ((k^2-6k+9)*(k^2+9))/((k^2+9)*(k^2-9))

tan(x)+1/cos(x) = (k^2-6k+9)/(k^2-9)

tan(x)+1/cos(x) = (k-3)(k-3)/(k+3)(k-3)

tan(x)+1/cos(x) = (k-3)/(k+3)

I am about 99% that I am wrong, but I hope this helped. :))

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I'd love to help. 

So the median is the middle number, or if there are 2, the average of the 2 middle number. 

72 72 72 72 74 74 74 74 79 79 79

So there are 11 numbers, so the number number will be 6th one. (11/2+0.5). 

72 72 72 72 74 74 74 74 79 79 79

1    2   3   4    5  6    7   8  9  10 11

The 6th number is 74, I hope this helped. :))

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