I honestly went by elimination so
Work:
Teacher = 4 Student = 2
A) 5 Teachers = 20 pizza 1 Student = 2 = 22
B) 4 Teachers= 16 Pizza 2 Student = 4 = 20
C) 3 Teachers = 12 Pizza 3 Student = 6 = 18
D) 2 Teachers = 8 Pizza 4 Student = 8 = 16
E) 1 Teacher = 4 Pizza 5 Student= 10 = 14
Answer
There are exactly 3 Teachers and 3 Students. (Answer C)
Given:
Divide 73.85 by 4.1.
To find:
The first step when applying properties of operations to divide 73.85 by 4.1.
Solution:
We have,
Divide 73.85 by 4.1 ![=\dfrac{73.83}{4.1}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B73.83%7D%7B4.1%7D)
Here, the dividend is 73.83 and divisor is 4.1.
First, we have to remove the decimals.
In numerator we have to digits after decimal and in denominator we have 1 digit after decimal.
So, multiply the divisor and dividend by 100 to remove the decimal.
Therefore, the first step is "Multiply the divisor and dividend by 100".
Answer:
Every one out of 125 people got a mild allergic reaction or 0.8%
Step-by-step explanation:
64/8000 simplified by 8 is 8/1000 simplified by 4 is 2/250 simplified by 2 is 1/125
Take the derivative with respect to t
![- w \sin(wt) + \sqrt{8} w cos(wt)](https://tex.z-dn.net/?f=%20-%20w%20%5Csin%28wt%29%20%2B%20%5Csqrt%7B8%7D%20w%20cos%28wt%29)
the maximum and minimum values occur when the tangent line is zero so we set the derivative to zero
![0 = -w \sin(wt) + \sqrt{8} w cos(wt)](https://tex.z-dn.net/?f=0%20%3D%20-w%20%5Csin%28wt%29%20%2B%20%5Csqrt%7B8%7D%20w%20cos%28wt%29)
divide by w
![0 =- \sin(wt) + \sqrt{8} cos(wt)](https://tex.z-dn.net/?f=0%20%3D-%20%5Csin%28wt%29%20%2B%20%5Csqrt%7B8%7D%20cos%28wt%29)
we add sin(wt) to both sides
![\sin(wt)= \sqrt{8} cos(wt)](https://tex.z-dn.net/?f=%5Csin%28wt%29%3D%20%5Csqrt%7B8%7D%20cos%28wt%29)
divide both sides by cos(wt)
![\frac{sin(wt)}{cos(wt)}= \sqrt{8} \\ \\ arctan(tan(wt))=arctan( \sqrt{8} ) \\ \\ wt=arctan(2 \sqrt{2)}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bsin%28wt%29%7D%7Bcos%28wt%29%7D%3D%20%5Csqrt%7B8%7D%20%20%20%5C%5C%20%20%5C%5C%20arctan%28tan%28wt%29%29%3Darctan%28%20%5Csqrt%7B8%7D%20%29%20%5C%5C%20%20%5C%5C%20wt%3Darctan%282%20%5Csqrt%7B2%29%7D%20)
OR
![wt=arctan( { \frac{1}{ \sqrt{2} } )](https://tex.z-dn.net/?f=wt%3Darctan%28%20%7B%20%5Cfrac%7B1%7D%7B%20%5Csqrt%7B2%7D%20%7D%20%29%20)
(wt)=2(n*pi-arctan(2^0.5))
(wt)=2(n*pi+arctan(2^-0.5))
where n is an integer
the absolute max and min will be
![I=cos(2n \pi -2arctan( \sqrt{2} ))](https://tex.z-dn.net/?f=I%3Dcos%282n%20%5Cpi%20-2arctan%28%20%5Csqrt%7B2%7D%20%29%29)
since 2npi is just the period of cos
![cos(2arctan( \sqrt{2} ))= \frac{-1}{3} ](https://tex.z-dn.net/?f=cos%282arctan%28%20%5Csqrt%7B2%7D%20%29%29%3D%20%5Cfrac%7B-1%7D%7B3%7D%20%0A)
substituting our second soultion we get
![I=cos(2n \pi +2arctan( \frac{1}{ \sqrt{2} } ))](https://tex.z-dn.net/?f=I%3Dcos%282n%20%5Cpi%20%2B2arctan%28%20%5Cfrac%7B1%7D%7B%20%5Csqrt%7B2%7D%20%7D%20%29%29)
since 2npi is the period
![I=cos(2arctan( \frac{1}{ \sqrt{2}} ))= \frac{1}{3}](https://tex.z-dn.net/?f=I%3Dcos%282arctan%28%20%5Cfrac%7B1%7D%7B%20%5Csqrt%7B2%7D%7D%20%29%29%3D%20%5Cfrac%7B1%7D%7B3%7D%20)
so the maximum value =
![\frac{1}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B3%7D%20)
minimum value =
Answer:5/12 (Decimal: 0.416667)
Step-by-step explanation: