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>
so
<u>the surface area of the box is equal to</u>
(remember that the box is open)
area of the base=
Perimeter of the base=
height=(h) m
substitute
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,∞)
The general equation of a hyperbola with a horizontal transverse axis is defined as:
x²/a² - y²/b² = 1
Solving for b², we use the formula: a² + b² = c²
b² = 12² - 9² = 63
Equation of our hyperbola will be:
x²/81 - y²/63 = 1
Answer:
so it can have a better balence
Step-by-step explanation:
-15x - 3y = 3
-15x - 3(5x-1) = 3 [input y]
-15x - 15x - 3 = 3 [distribute 3 over 5x and -1]
-30x - 3 = 3 [combine like terms]
-30x = 6 [add 3 to eliminate the negative]
x =
[divide by -30 to get x alone]
x =
[reduce the fraction]
Hopefully this is helpful enough to get you to your answer
