Answer: x = -5, x = 6
<u>Step-by-step explanation:</u>
"Solutions" are also called roots, zeroes, and x-intercepts and are where the parabola crosses the x-axis.
The parabola crosses the x-axis at x = -5 and at x = 6
(see attachment)
Y=-2x-3 thats the final answer
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>
Hello!
The answer is: 15 mL
Why?
From the statement we have that 250 mg need to be given every 8 hours, we can calculate how many mg will be given each day (24 hours):
![\frac{24Hours}{8Hours} = 3](https://tex.z-dn.net/?f=%5Cfrac%7B24Hours%7D%7B8Hours%7D%20%3D%203)
So,
a day
We also know that there are 50 mg cefaclor every 1mL,
So, to know how many mL will be given with 750mg, we just need to divide it by 50
![\frac{750}{50} = 15](https://tex.z-dn.net/?f=%5Cfrac%7B750%7D%7B50%7D%20%3D%2015)
Therefore,
15mL or 750 mg of cefaclor will be given each day.
Have a nice day!
(x - 1)2 - 4(X + 1) + 2 = 0
2x - 2 - 4x - 4 + 2 = 0
(2x - 4x) - 4 = 0
-2x = 4
x = 4/-2
x = -2