Let
x--------> the length side of the square base
h--------> the height of the box
we know that
<u>the volume of the box is equal to</u>
![V=x^{2} *h\\ V=9\ m^{3}](https://tex.z-dn.net/?f=V%3Dx%5E%7B2%7D%20%2Ah%5C%5C%20V%3D9%5C%20m%5E%7B3%7D)
so
![x^{2} *h=9\\\\h=\frac{9}{x^{2} }](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2Ah%3D9%5C%5C%5C%5Ch%3D%5Cfrac%7B9%7D%7Bx%5E%7B2%7D%20%7D)
<u>the surface area of the box is equal to</u>
(remember that the box is open)
area of the base=![(x^{2})\ m^{2}](https://tex.z-dn.net/?f=%28x%5E%7B2%7D%29%5C%20m%5E%7B2%7D)
Perimeter of the base=![(4*x)\ m](https://tex.z-dn.net/?f=%284%2Ax%29%5C%20m)
height=(h) m
![h=\frac{9}{x^{2}}](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B9%7D%7Bx%5E%7B2%7D%7D)
substitute
![SA=x^{2} +4*x*h\\ \\ \\ SA=x^{2} +4*x*\frac{9}{x^{2} } \\ \\ SA=x^{2} +\frac{36}{x}](https://tex.z-dn.net/?f=SA%3Dx%5E%7B2%7D%20%2B4%2Ax%2Ah%5C%5C%20%5C%5C%20%5C%5C%20SA%3Dx%5E%7B2%7D%20%2B4%2Ax%2A%5Cfrac%7B9%7D%7Bx%5E%7B2%7D%20%7D%20%5C%5C%20%5C%5C%20SA%3Dx%5E%7B2%7D%20%2B%5Cfrac%7B36%7D%7Bx%7D)
we know that
the value of x can not be negative and the denominator can not be zero
therefore
<u>the answer is</u>
the domain of SA is x> 0
the domain is the interval-------------> (0,∞)
0 there are no odd number 1-15
Answer: I do not think that failing one class will effect your entire 6th grade year. Although, this is best discussed with your teachers or parents/guardians.
I believe the answer is y = -5/2x + 6
![4x^2+9y^2=36\\ \\ \frac{4x^2}{36}+\frac{9y^2}{36}=\frac{36}{36}\\ \\ \boxed{\frac{x^2}{9}+\frac{y^2}{4}=1}](https://tex.z-dn.net/?f=4x%5E2%2B9y%5E2%3D36%5C%5C%0A%5C%5C%0A%5Cfrac%7B4x%5E2%7D%7B36%7D%2B%5Cfrac%7B9y%5E2%7D%7B36%7D%3D%5Cfrac%7B36%7D%7B36%7D%5C%5C%0A%5C%5C%0A%5Cboxed%7B%5Cfrac%7Bx%5E2%7D%7B9%7D%2B%5Cfrac%7By%5E2%7D%7B4%7D%3D1%7D)
This is a equation of a ellipse (0,0) centered
Domais: {x∈R/-3≤x≤3}
Range:{y∈R/-2≤y≤2}