To calculate the inverse isolate x on one side and switch x with y on both sides
1)
y=(2x-3)/5
5y=2x-3
5y+3=2x
(5y+3)/2=x
-> f(x)=(5x+3)/2
2)
y=(x+8)/2
2y=x+8
2y-8=x
-> f(x)=2x-8
3)
y=(x+2)/7
7y=x+2
7y-2=x
-> f(x)=7x-2
4)
y=(1-2x)/x
y=(1/x)-2
y+2=1/x
x=1/(y+2)
-> f(x)=1/(x+2)
Answer:
6.9
7.-5
8.-10
Step-by-step explanation:
Answer:
The cat should receive 1/5 of the can of food at each meal.
Step-by-step explanation:
Given that a cat is to receive 90 kcal of a canned food at each meal, and the food has a caloric density of 450 kcal per can, to determine what fraction of a can it should receive at each meal, the following calculation must be performed:
1 / (450/90) = X
1/5 = X
Thus, the cat should receive 1/5 of the can of food at each meal.
Answer:
see below
Step-by-step explanation:
Dosage= 500 mg
Frequency= twice a day (every 12 hours)
Duration= 10 days
Number of dosage= 10*2= 20
residual drug amount after each dosage= 4.5%
We can build an equation to calculate residual drug amount:
d= 500*(4.5/100)*t= 22.5t, where d- is residual drug, t is number of dosage
After first dose residual drug amount is:
After second dose:
As per the equation, the higher the t, the greater the residual drug amount in the body.
Maximum residual drug will be in the body:
- d= 20*22.5= 450 mg at the end of 10 days
Maximum drug will be in the body right after the last dose, when the amount will be:
