Find modulus and principle argument of compiles numbers Z=5/2(cos300+isin300)
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Option B. y = 3 (one-third) Superscript x is the function that is graphed in this question.
<h3>How to solve the problem</h3>We have exponential functions to be of the form abˣ
This can be written in the form of f(x) =
So it crosses through the point (0, 3).
From the way that the curve passes through the circle, we can see that the it is said to pass through at (0, 3) and approaches y = 0 in quadrant 1
Hence this would put our answer to be y = 3 (one-third) Superscript x
Read more on coordinate planes here
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Let h = height of the box,
x = side length of the base.
Volume of the box is
.
So
Surface area of a box is S = 2(Width • Length + Length • Height + Height • Width).
So surface area of the box is
The surface are is supposed to be the minimum. So we'll need to find the first derivative of the surface area function and set it to zero.
![4x = \frac{460}{ x^{2} } \\ 4x^{3} = 460 \\ x^{3} = 115 \\ x = \sqrt[3]{115} = 4.86](https://tex.z-dn.net/?f=%204x%20%3D%20%5Cfrac%7B460%7D%7B%20x%5E%7B2%7D%20%7D%20%20%5C%5C%20%204x%5E%7B3%7D%20%3D%20460%20%20%5C%5C%20x%5E%7B3%7D%20%3D%20115%20%20%5C%5C%20x%20%3D%20%20%5Csqrt%5B3%5D%7B115%7D%20%3D%204.86%20)
Then
So the box is 4.86 in. wide and 4.87 in. high.
x = side length of the base.
Volume of the box is
So
Surface area of a box is S = 2(Width • Length + Length • Height + Height • Width).
So surface area of the box is
The surface are is supposed to be the minimum. So we'll need to find the first derivative of the surface area function and set it to zero.
Then
So the box is 4.86 in. wide and 4.87 in. high.