Answer:
54 teachers
Step-by-step explanation:
Step one:
given data
The ratio of teachers to students on a field trip to New York City is 2 to 9.
That is the ratio is 2:9
We are told that there are 243 students
Required
We want to find the number of teachers
Step two:
Applying the part to part method we have
let the number of teachers be x
2/9= x/243
cross multiply we have
9x= 243*2
9x= 486
divide both sides by 9
x= 54
Hence there are 54 teachers
12/16 are shaded. 12/16 simplified is 3/4 or 75%.
I think its probaly going to be 6
Answer:
x = 8
Step-by-step explanation:
Firstly, create an equation or formula
- because they are congurent, IN = AT, so:
2x+10 = 26 (minus 10 from both sides)
2x = 16 (divide both sides by 2)
x = 8
Sub into the formula:
2 x 8 + 10 = 26
26 = 26
Answer: The required solution is 
Step-by-step explanation:
We are given to solve the following differential equation :

where k is a constant and the equation satisfies the conditions y(0) = 50, y(5) = 100.
From equation (i), we have

Integrating both sides, we get
![\int\dfrac{dy}{y}=\int kdt\\\\\Rightarrow \log y=kt+c~~~~~~[\textup{c is a constant of integration}]\\\\\Rightarrow y=e^{kt+c}\\\\\Rightarrow y=ae^{kt}~~~~[\textup{where }a=e^c\textup{ is another constant}]](https://tex.z-dn.net/?f=%5Cint%5Cdfrac%7Bdy%7D%7By%7D%3D%5Cint%20kdt%5C%5C%5C%5C%5CRightarrow%20%5Clog%20y%3Dkt%2Bc~~~~~~%5B%5Ctextup%7Bc%20is%20a%20constant%20of%20integration%7D%5D%5C%5C%5C%5C%5CRightarrow%20y%3De%5E%7Bkt%2Bc%7D%5C%5C%5C%5C%5CRightarrow%20y%3Dae%5E%7Bkt%7D~~~~%5B%5Ctextup%7Bwhere%20%7Da%3De%5Ec%5Ctextup%7B%20is%20another%20constant%7D%5D)
Also, the conditions are

and

Thus, the required solution is 