- f-¹(x) = 69/5y
- f-¹(x) = 13 + 4y/(y + 10)
<h3>How to determine the inverse </h3>
The steps involved in determining the inverse of a function are;
- Replace f(x) with y in the equation describing the function
- Interchange x and y
- Solve for y
- Replace y by f-1(x)
If f(x) = 4/5x + 13
y = 4/5y + 13
y = 4 + 65 /5y
f^-1 = 4 + 65 /5y
f^-1(x)= 69/5y
If f(x)= 3/x+10 + 4
y = 3/y + 10 + 4
y = 3 + 4y + 10/(y + 10)
y = 13 + 4y/(y + 10)
f^-1(x) = 13 + 4y/(y + 10)
Thus, the inverse of the functions is 69/5y and 13 + 4y/(y + 10)
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Answer:
The missing length would be 24.
First, you find the ratio of 55 to 33.
This gets you approximately 1.66666667
Multiply 36 by 1.66666667. This will get you 60.
60 - 36 = 24
The unknown value is 24.
Answer:
D. 80
Step-by-step explanation:
multiply 40 by 2 cause its 50 percent
Answer:
If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is is dA/dt = 15 - 0.005A
Option C) dA/dt = 15 - 0.005A is the correction Answer
Step-by-step explanation:
Given the data in the question;
If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is?
dA/dt = rate in - rate out
first we determine the rate in and rate out;
rate in = 3pound/gallon × 5gallons/min = 15 pound/min
rate out = A pounds/1000gallons × 5gallons/min = 5Ag/1000pounds/min
= 0.005A pounds/min
so we substitute
dA/dt = rate in - rate out
dA/dt = 15 - 0.005A
Therefore, If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is is dA/dt = 15 - 0.005A
Option C) dA/dt = 15 - 0.005A is the correction Answer
here is your ans have a nice day
