Hello! And thank you for your question!
First add 7 to both sides:
x/5 = 4 + 7
Then simplify 4 + 7:
x/5 = 11
After that, multiply both sides by 5
11 x 5 = x
Finally, Simplify 11 * 5:
x = 55
Final Answer:
x = 55
Answer:
It should be the last one, She made a sign error when multiplying.
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.
Since the x value can be changed to anything other than 4 (because then you would be dividing by 0 in x-4), the answer is: All real numbers except 4.
Answer:

Step-by-step explanation:
We are given that a differential equation

We have to find the general solution of given differential equation


Compare with

We get


I.F=




