Hey there!
One way to do this is find all the factors of 6 and then see which pair fit the requirements.
The factors of 6 are 1, 2, 3, and 6. (Note: There can be negative factors, but I am going to leave them out since it is asking for positive integers.)
You can find them by asking if each number can go into 6.
1, 2, 3, and 6 all go into 6, while 4 and 5 do not.
The requirements we have is that they must be consecutive <u>and</u> have a product of 6.
Consecutive means right after one another.
The only numbers that fit this are 2 and 3.
2 x 3 = 6
Hope this helps!
there are many combinations for it, but we can settle for say
![\bf \begin{cases} f(x)=x+2\\[1em] g(x)=\cfrac{9}{x^2}\\[-0.5em] \hrulefill\\ (f\circ g)(x)\implies f(~~g(x)~~) \end{cases}\qquad \qquad f(~~g(x)~~)=[g(x)]+2 \\\\\\ f(~~g(x)~~)=\left[ \cfrac{9}{x^2} \right]+2\implies f(~~g(x)~~)=\cfrac{9}{x^2}+2](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20f%28x%29%3Dx%2B2%5C%5C%5B1em%5D%20g%28x%29%3D%5Ccfrac%7B9%7D%7Bx%5E2%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20%28f%5Ccirc%20g%29%28x%29%5Cimplies%20f%28~~g%28x%29~~%29%20%5Cend%7Bcases%7D%5Cqquad%20%5Cqquad%20f%28~~g%28x%29~~%29%3D%5Bg%28x%29%5D%2B2%20%5C%5C%5C%5C%5C%5C%20f%28~~g%28x%29~~%29%3D%5Cleft%5B%20%5Ccfrac%7B9%7D%7Bx%5E2%7D%20%5Cright%5D%2B2%5Cimplies%20f%28~~g%28x%29~~%29%3D%5Ccfrac%7B9%7D%7Bx%5E2%7D%2B2)
Answer: Number one would be 3x+2y+-14 and x+y=-4, second one is x=-3, y=7
Step-by-step explanation:
The first one is solved by inputting the x and y in each one and finding which one comes out true, the second on is solved by substitution. to find x you would subtract x in the first equation and make it y=4-x then input that equation in the y in the second equation.
3x1000+2x100+4x10+6x1+7x(1/10)+8x(1/100)
Sorry but I don’t understand your language.