The y-intercept is (0, -12).
Hope that helped. :)
Answer:
The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches
Step-by-step explanation:
We have that:
![Area = 128](https://tex.z-dn.net/?f=Area%20%3D%20128)
Let the dimension of the paper be x and y;
Such that:
![Length = x](https://tex.z-dn.net/?f=Length%20%3D%20x)
![Width = y](https://tex.z-dn.net/?f=Width%20%3D%20y)
So:
![Area = x * y](https://tex.z-dn.net/?f=Area%20%3D%20x%20%2A%20y)
Substitute 128 for Area
![128 = x * y](https://tex.z-dn.net/?f=128%20%3D%20x%20%2A%20y)
Make x the subject
![x = \frac{128}{y}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B128%7D%7By%7D)
When 1 inch margin is at top and bottom
The length becomes:
![Length = x + 1 + 1](https://tex.z-dn.net/?f=Length%20%3D%20x%20%2B%201%20%2B%201)
![Length = x + 2](https://tex.z-dn.net/?f=Length%20%3D%20x%20%2B%202)
When 2 inch margin is at both sides
The width becomes:
![Width = y + 2 + 2](https://tex.z-dn.net/?f=Width%20%3D%20y%20%2B%202%20%2B%202)
![Width = y + 4](https://tex.z-dn.net/?f=Width%20%3D%20y%20%2B%204)
The New Area (A) is then calculated as:
![A = (x + 2) * (y + 4)](https://tex.z-dn.net/?f=A%20%3D%20%28x%20%2B%202%29%20%2A%20%28y%20%2B%204%29)
Substitute
for x
![A = (\frac{128}{y} + 2) * (y + 4)](https://tex.z-dn.net/?f=A%20%3D%20%28%5Cfrac%7B128%7D%7By%7D%20%2B%202%29%20%2A%20%28y%20%2B%204%29)
Open Brackets
![A = 128 + \frac{512}{y} + 2y + 8](https://tex.z-dn.net/?f=A%20%3D%20128%20%2B%20%5Cfrac%7B512%7D%7By%7D%20%2B%202y%20%2B%208)
Collect Like Terms
![A = \frac{512}{y} + 2y + 8+128](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7B512%7D%7By%7D%20%2B%202y%20%2B%208%2B128)
![A = \frac{512}{y} + 2y + 136](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7B512%7D%7By%7D%20%2B%202y%20%2B%20136)
![A= 512y^{-1} + 2y + 136](https://tex.z-dn.net/?f=A%3D%20512y%5E%7B-1%7D%20%2B%202y%20%2B%20136)
To calculate the smallest possible value of y, we have to apply calculus.
Different A with respect to y
![A' = -512y^{-2} + 2](https://tex.z-dn.net/?f=A%27%20%3D%20-512y%5E%7B-2%7D%20%2B%202)
Set
![A' = 0](https://tex.z-dn.net/?f=A%27%20%3D%200)
This gives:
![0 = -512y^{-2} + 2](https://tex.z-dn.net/?f=0%20%3D%20-512y%5E%7B-2%7D%20%2B%202)
Collect Like Terms
![512y^{-2} = 2](https://tex.z-dn.net/?f=512y%5E%7B-2%7D%20%3D%202)
Multiply through by ![y^2](https://tex.z-dn.net/?f=y%5E2)
![y^2 * 512y^{-2} = 2 * y^2](https://tex.z-dn.net/?f=y%5E2%20%2A%20512y%5E%7B-2%7D%20%3D%202%20%2A%20y%5E2)
![512 = 2y^2](https://tex.z-dn.net/?f=512%20%3D%202y%5E2)
Divide through by 2
![256=y^2](https://tex.z-dn.net/?f=256%3Dy%5E2)
Take square roots of both sides
![\sqrt{256=y^2](https://tex.z-dn.net/?f=%5Csqrt%7B256%3Dy%5E2)
![16=y](https://tex.z-dn.net/?f=16%3Dy)
![y = 16](https://tex.z-dn.net/?f=y%20%3D%2016)
Recall that:
![x = \frac{128}{y}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B128%7D%7By%7D)
![x = \frac{128}{16}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B128%7D%7B16%7D)
![x = 8](https://tex.z-dn.net/?f=x%20%3D%208)
Recall that the new dimensions are:
![Length = x + 2](https://tex.z-dn.net/?f=Length%20%3D%20x%20%2B%202)
![Width = y + 4](https://tex.z-dn.net/?f=Width%20%3D%20y%20%2B%204)
So:
![Length = 8 + 2](https://tex.z-dn.net/?f=Length%20%3D%208%20%2B%202)
![Length = 10](https://tex.z-dn.net/?f=Length%20%3D%2010)
![Width = 16 + 4](https://tex.z-dn.net/?f=Width%20%3D%2016%20%2B%204)
![Width = 20](https://tex.z-dn.net/?f=Width%20%3D%2020)
To double-check;
Differentiate A'
![A' = -512y^{-2} + 2](https://tex.z-dn.net/?f=A%27%20%3D%20-512y%5E%7B-2%7D%20%2B%202)
![A" = -2 * -512y^{-3}](https://tex.z-dn.net/?f=A%22%20%3D%20-2%20%2A%20-512y%5E%7B-3%7D)
![A" = 1024y^{-3}](https://tex.z-dn.net/?f=A%22%20%3D%201024y%5E%7B-3%7D)
![A" = \frac{1024}{y^3}](https://tex.z-dn.net/?f=A%22%20%3D%20%5Cfrac%7B1024%7D%7By%5E3%7D)
The above value is:
![A" = \frac{1024}{y^3} > 0](https://tex.z-dn.net/?f=A%22%20%3D%20%5Cfrac%7B1024%7D%7By%5E3%7D%20%3E%200)
This means that the calculated values are at minimum.
<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>
Answer:
I believe the answer is 12x 5
Tan(angle) = opposite/adjacent
tan(x) = 90/51
x = arctan(90/51)
x = 60.461217740442
which rounds to 60 when rounding to the nearest whole number
Answer: 60
Answer:
tan A = 5/12
Step-by-step explanation:
To find tangent of a triangle, it is this ratio : opposite / adjacent.
Knowing that "A" is the given angle, 5 is the opposite and 12 is the side adjacent to "A". Putting these values together, the tangent is
5 (opposite)/12 (adjacent).
Hope this helps!