Melody

Two numbers, if the first one increases by 1, and the second one decreases by 1, then their product increases by 2020. If the first number decreases by 7, and the second one increases by 7, what value does the product decrease?

let the numbers be x and y   and let the product decrease by k

\((x+1)(y-1)=xy+2020\\ xy-x+y-1=xy+2020\\ -x+y-1=2020\\ y-x=2021\qquad (1) \)
 

\((x-7)(y+7)=xy - k\\ xy+7x-7y-49=xy - k\\ 7x-7y-49= - k\\ 7(x-y)-49=-k\\ k=49-7(x-y)\\ k=49-7*-2021\\ k=49+7*2021\\ k=14196\)          

it is up to the asker to check that the answer is correct.

atan stands for arctan     and     arctan is the same as inverse tan.

so

tan^(-1)

or

atan      

or

you can use the  x²  button in the ribbon and have    tan^-1

or

you can use Latex. Open the Latex box in the ribbon and type  tan^{-1}   and you get      \(tan^{-1}\)

A positive integer k is called a "lucky" number if the units digit of k^6 is 4. How many lucky numbers smaller than 1000 are there?

For this to happen the units value of k must be 2 or 8
so, 2,8,12,18, etc

1000/10 *2 = 200

there will be 200 lucky values of k

\(g^4 + 12g^2 + 9 - 6g^2 + 7\\ [g^4 + 12g^2+(\frac{12}{2})^2 ]+ 9 - 6g^2 + 7-(\frac{12}{2})^2\)

You can take it from there.

reposted here:

This is a repost of this one

\(531n \equiv 1067n \pmod{24}\\ 24*22n+3n \equiv 24*44n+11n \pmod{24}\\ 0+3n \equiv 0+11n \pmod{24}\\ 3n \equiv 11n \pmod{24}\\ 8n \equiv 0 \pmod{24}\\ n=0,3,6,9,etc\\ \text{The smallest positive n is 3}\)

f(x) = x^2 - 3x + 4

f(2x+1) = (2x+1)^2 - 3(2x+1) + 4

now simplify

y = C1e^−3x + C2e^−2x + C3e^−x

do you mean

y = C_1e^(−3x) + C_2e^(−2x) + C_3e^(−x)

I do not know what you want me to do with this but I could rearrange it as follows.

\(y = C_1e^{−3x} + C_2e^{−2x} + C_3e^{−x}\\~\\ y = e^{-3x}(C_1 + C_2e^{x} + C_3e^{2x})\\~\\ y = \frac{C_1 + C_2e^{x} + C_3e^{2x}}{e^{3x}}\\~\\ \)