Answer:
<h2>7</h2>
Step-by-step explanation:
![\left[\left(11\:-\:4\right)^3\right]^2\:\div \left(4\:+\:3\right)^5\\\\\frac{\left(\left(11-4\right)^3\right)^2}{\left(4+3\right)^5}\\\\\mathrm{Subtract\:the\:numbers:}\:11-4=7\\\\=\frac{\left(7^3\right)^2}{\left(4+3\right)^5}\\\\\mathrm{Add\:the\:numbers:}\:4+3=7\\\\=\frac{\left(7^3\right)^2}{7^5}\\\\\left(7^3\right)^2=7^6\\\\=\frac{7^6}{7^5}\\\\\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}=x^{a-b}\\\\\frac{7^6}{7^5}=7^{6-5}\\\\\mathrm{Subtract\:the\:numbers:}\:6-5=1\\\\=7](https://tex.z-dn.net/?f=%5Cleft%5B%5Cleft%2811%5C%3A-%5C%3A4%5Cright%29%5E3%5Cright%5D%5E2%5C%3A%5Cdiv%20%5Cleft%284%5C%3A%2B%5C%3A3%5Cright%29%5E5%5C%5C%5C%5C%5Cfrac%7B%5Cleft%28%5Cleft%2811-4%5Cright%29%5E3%5Cright%29%5E2%7D%7B%5Cleft%284%2B3%5Cright%29%5E5%7D%5C%5C%5C%5C%5Cmathrm%7BSubtract%5C%3Athe%5C%3Anumbers%3A%7D%5C%3A11-4%3D7%5C%5C%5C%5C%3D%5Cfrac%7B%5Cleft%287%5E3%5Cright%29%5E2%7D%7B%5Cleft%284%2B3%5Cright%29%5E5%7D%5C%5C%5C%5C%5Cmathrm%7BAdd%5C%3Athe%5C%3Anumbers%3A%7D%5C%3A4%2B3%3D7%5C%5C%5C%5C%3D%5Cfrac%7B%5Cleft%287%5E3%5Cright%29%5E2%7D%7B7%5E5%7D%5C%5C%5C%5C%5Cleft%287%5E3%5Cright%29%5E2%3D7%5E6%5C%5C%5C%5C%3D%5Cfrac%7B7%5E6%7D%7B7%5E5%7D%5C%5C%5C%5C%5Cmathrm%7BApply%5C%3Aexponent%5C%3Arule%7D%3A%5Cquad%20%5Cfrac%7Bx%5Ea%7D%7Bx%5Eb%7D%3Dx%5E%7Ba-b%7D%5C%5C%5C%5C%5Cfrac%7B7%5E6%7D%7B7%5E5%7D%3D7%5E%7B6-5%7D%5C%5C%5C%5C%5Cmathrm%7BSubtract%5C%3Athe%5C%3Anumbers%3A%7D%5C%3A6-5%3D1%5C%5C%5C%5C%3D7)
<h3>
Answer: x = (
y-2)^2 +
5</h3>
In other words, y-2 goes in the first box and 5 goes in the second box.
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Work Shown:
y^2 - 4y - x + 9 = 0
y^2 - 4y + 9 = x
x = y^2 - 4y + 9
x = y^2 - 4y + 4 + 5 .... rewrite 9 as 4+5
x = (y^2-4y+4) + 5
x = (y-2)^2 + 5 .... apply the perfect square factoring rule
So we'll have y-2 go in the first box and 5 goes in the second box
note: One version of the perfect square factoring rule says (a-b)^2 = a^2-2ab+b^2.
Less than sign

Greater than sign

Less than or equal to sign

Greater than or equal to sign