Given:
Volume of cuboid container = 2 litres
The container has a square base.
Its height is double the length of each edge on its base.
To find:
The height of the container.
Solution:
We know that,
1 litre = 1000 cubic cm
2 litre = 2000 cubic cm
Let x be the length of each edge on its base. Then the height of the container is:
The volume of a cuboid is:
Where, l is length, w is width and h is height.
Putting 
, we get
Divide both sides by 2.
Taking cube root on both sides.
![\sqrt[3]{1000}=x](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B1000%7D%3Dx)
Now, the height of the container is:
Therefore, the height of the container is 20 cm.
A.) Simplifying the equation would lead you to 15x^2-3x+9
b.) You know your answer is correct because you're adding the two polynomials together. 9x^2+6x^2 is 1tx^2. Now you have 2x-5x and since the negative is bigger, you get -3x. Then 5+4 is 9. You have no like terms therefore your answer is 15x^2-3x+9
The surface area would be 68 squared inches. I hope this helps!
We are given the following:
- parabola passes to both (1,0) and (0,1)
<span> - slope at x = 1 is 4 from the equation of the tangent line </span>
<span>First, we figure out the value of c or the y intercept, we use the second point (0, 1) and substitute to the equation of the parabola. W</span><span>hen x = 0, y = 1. So, c should be equal to 1. The</span><span> parabola is y = ax^2 + bx + 1 </span>
<span>Now, we can substitute the point (1,0) into the equation,
</span>0 = a(1)^2 + b(1) + 1
<span>0 = a + b + 1
a + b = -1 </span>
<span>The slope at x = 1 is equal to 4 which is equal to the first derivative of the equation.</span>
<span>We take the derivative of the equation ,
y = ax^2 + bx + 1</span>
<span>y' = 2ax + b
</span>
<span>x = 1, y' = 2
</span>4 = 2a(1) + b
<span>4 = 2a + b </span>
So, we have two equations and two unknowns,<span> </span>
<span>2a + b = 4 </span>
<span>a + b = -1
</span><span>
Solving simultaneously,
a = 5 </span>
<span>b = -6</span>
<span>Therefore, the eqution of the parabola is y = 5x^2 - 6x + 1 .</span>
