Answer:
- Base Length of 84cm
- Height of 42 cm.
Step-by-step explanation:
Given a box with a square base and an open top which must have a volume of 296352 cubic centimetre. We want to minimize the amount of material used.
Step 1:
Let the side length of the base =x
Let the height of the box =h
Since the box has a square base
Volume, 

Surface Area of the box = Base Area + Area of 4 sides

Step 2: Find the derivative of A(x)

Step 3: Set A'(x)=0 and solve for x
![A'(x)=\dfrac{2x^3-1185408}{x^2}=0\\2x^3-1185408=0\\2x^3=1185408\\$Divide both sides by 2\\x^3=592704\\$Take the cube root of both sides\\x=\sqrt[3]{592704}\\x=84](https://tex.z-dn.net/?f=A%27%28x%29%3D%5Cdfrac%7B2x%5E3-1185408%7D%7Bx%5E2%7D%3D0%5C%5C2x%5E3-1185408%3D0%5C%5C2x%5E3%3D1185408%5C%5C%24Divide%20both%20sides%20by%202%5C%5Cx%5E3%3D592704%5C%5C%24Take%20the%20cube%20root%20of%20both%20sides%5C%5Cx%3D%5Csqrt%5B3%5D%7B592704%7D%5C%5Cx%3D84)
Step 4: Verify that x=84 is a minimum value
We use the second derivative test

Since the second derivative is positive at x=84, then it is a minimum point.
Recall:

Therefore, the dimensions that minimizes the box surface area are:
- Base Length of 84cm
- Height of 42 cm.





Consider a
ABC right angled at C and
Then,
‣ Base [B] = BC
‣ Perpendicular [P] = AC
‣ Hypotenuse [H] = AB

Let,
Base = 7k and Perpendicular = 8k, where k is any positive integer
In
ABC, H² = B² + P² by Pythagoras theorem






Calculating Sin




Calculating Cos




<u>Solving the given expression</u><u> </u><u>:</u><u>-</u><u> </u>

Putting,
• Sin
= 
• Cos
= 

<u>Using</u><u> </u><u>(</u><u>a</u><u> </u><u>+</u><u> </u><u>b</u><u> </u><u>)</u><u> </u><u>(</u><u>a</u><u> </u><u>-</u><u> </u><u>b</u><u> </u><u>)</u><u> </u><u>=</u><u> </u><u>a²</u><u> </u><u>-</u><u> </u><u>b²</u>










✧ Basic Formulas of Trigonometry is given by :-


✧ Figure in attachment

Answer:
William was 0.5 points away from a perfect score.
Step-by-step explanation:
If the teacher didn't subtract 5 points, he would have 24.5 + 5 = 29.5 points.
30 - 29.5 = 0.5
William was 0.5 points away from a perfect score.
See attached scatter plot
this does not show a linear pattern