Answer:
Step-by-step explanation:
Given
Suppose side face have a dimension of l\times l
and width of h
volume 
volume 
Area of side wall is 
Area of top Wall 
Area of bottom 
Cost of bottom wall 
Cost of top wall 
Cost of side walls 
total cost 
differentiate C w.r.t l to get minima or maxima
Dimension of Box is
Answer:
D its really basic math you can also just do it in a calculator lel
Answer:
It will take Shanti 150 square inches of paper to wrap the gifts
Step-by-step explanation:
Shanti wraps the gift she is taking to a friend's birthday party. She doesn't have very much paper, so she does not overlap any of the edges. The side lengths of the cube are all 5in.
How much paper did it take to cover the gift?
Solution
To solve this, you will find the surface area of a cube
Surface area of a cube = 6a^2
Where,
a=side length of the cube
a= 5 in
a^2=a × a
=5 × 5
=25 in^2
a^2=25 in^2
Surface area of a cube = 6a^2
=6 × 25 in^2
=150 in^2
It will take Shanti 150 square inches of paper to wrap the gifts
The differential equation that models the given situation will be dy/dt = K(N - y).
<h3>How to compute the equation?</h3>
Let y = number of individuals who have heard about the product.
Let N - y = this who haven't heard about the product.
From the given statement, the rate for change will be t = dy/dt. Therefore, the differential equation that models the given situation will be dy/dt = K(N - y).
Learn more about equations on:
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Answer:
![r = \sqrt[3]{\frac{3V}{4 \pi}}](https://tex.z-dn.net/?f=r%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B3V%7D%7B4%20%5Cpi%7D%7D)
Step-by-step explanation:
From the formula of volume of a sphere we have to isolate "r" on one side of the equation i.e. we have to make "r" the subject of the equation.
![V=\frac{4}{3} \pi r^{3}\\\\ \text{Multiplying both sides by 3/4 we get}\\\\\frac{3V}{4} = \pi r^{3}\\\\ \text{Dividing both sides by } \pi \\\\ \frac{3V}{4 \pi} = r^{3}\\\\\text{Takeing cube root of both sides}\\\\\sqrt[3]{\frac{3V}{4 \pi}} = r](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20r%5E%7B3%7D%5C%5C%5C%5C%20%5Ctext%7BMultiplying%20both%20sides%20by%203%2F4%20we%20get%7D%5C%5C%5C%5C%5Cfrac%7B3V%7D%7B4%7D%20%3D%20%5Cpi%20r%5E%7B3%7D%5C%5C%5C%5C%20%5Ctext%7BDividing%20both%20sides%20by%20%7D%20%5Cpi%20%5C%5C%5C%5C%20%5Cfrac%7B3V%7D%7B4%20%5Cpi%7D%20%3D%20r%5E%7B3%7D%5C%5C%5C%5C%5Ctext%7BTakeing%20cube%20root%20of%20both%20sides%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B%5Cfrac%7B3V%7D%7B4%20%5Cpi%7D%7D%20%3D%20r)
Therefore:
