Answer:
I can't understand what you have written and what to find
![\displaystyle (x+y)^n=\sum_{k=0}^n\binom{n}{k}x^{n-k}y^k](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%0A%28x%2By%29%5En%3D%5Csum_%7Bk%3D0%7D%5En%5Cbinom%7Bn%7D%7Bk%7Dx%5E%7Bn-k%7Dy%5Ek%20)
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![\displaystyle (x^2+y)^n=\sum_{k=0}^n\binom{n}{k}x^{2n-2k}y^k\\ n=5\\ k=3\\\\\binom{5}{3}=\dfrac{5!}{3!2!}=\dfrac{4\cdor5}{2}=10](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%0A%28x%5E2%2By%29%5En%3D%5Csum_%7Bk%3D0%7D%5En%5Cbinom%7Bn%7D%7Bk%7Dx%5E%7B2n-2k%7Dy%5Ek%5C%5C%0An%3D5%5C%5C%0Ak%3D3%5C%5C%5C%5C%5Cbinom%7B5%7D%7B3%7D%3D%5Cdfrac%7B5%21%7D%7B3%212%21%7D%3D%5Cdfrac%7B4%5Ccdor5%7D%7B2%7D%3D10%20)
<u>It's 10.</u>
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![\displaystyle (3x^2+y)^n=\sum_{k=0}^n\binom{n}{k}(3x)^{2n-2k}y^k=\sum_{k=0}^n\binom{n}{k}\cdot 3^{2n-2k}\cdot x^{2n-2k}y^k\\\\ n=5\\ k=4\\\\ \binom{5}{3}\cdot3^{2\cdot5-2\cdot4}=10\cdot3^{2}=10\cdot9=90](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%0A%283x%5E2%2By%29%5En%3D%5Csum_%7Bk%3D0%7D%5En%5Cbinom%7Bn%7D%7Bk%7D%283x%29%5E%7B2n-2k%7Dy%5Ek%3D%5Csum_%7Bk%3D0%7D%5En%5Cbinom%7Bn%7D%7Bk%7D%5Ccdot%203%5E%7B2n-2k%7D%5Ccdot%20x%5E%7B2n-2k%7Dy%5Ek%5C%5C%5C%5C%0An%3D5%5C%5C%0Ak%3D4%5C%5C%5C%5C%0A%5Cbinom%7B5%7D%7B3%7D%5Ccdot3%5E%7B2%5Ccdot5-2%5Ccdot4%7D%3D10%5Ccdot3%5E%7B2%7D%3D10%5Ccdot9%3D90)
<u>It's 90</u>
Your answer is: 20
They have 4 different backgrounds and 5 different poses which means you just have to multiply the 4 backgrounds*the 5 different poses to discover how many different pictures you could have at the most.
hope this helps!
3x-11=17-4x add 4x to both sides
7x-11=17 add 11 to both sides
7x=28 divide both sides by 7
x=4