Answer:
Patient A - 150 mg
Patient B - 50 mg
Step-by-step explanation:
Since patient A gets three times more, 50 times 3 is 150. 150 plus 50 is 200, therefore, patient A gets 150 mg and patient B gets 50 mg. I hope this helps :)
Y=600000(1+1.8%)^(x-2000)
So the population in 2012
Y=600,000×(1+0.018)^(2,012−2,000)
Y=743,232
Hello,
Here is the demonstration in the book Person Guide to Mathematic by Khattar Dinesh.
Let's assume
P=cos(a)*cos(2a)*cos(3a)*....*cos(998a)*cos(999a)
Q=sin(a)*sin(2a)*sin(3a)*....*sin(998a)*sin(999a)
As sin x *cos x=sin (2x) /2
P*Q=1/2*sin(2a)*1/2sin(4a)*1/2*sin(6a)*....
*1/2* sin(2*998a)*1/2*sin(2*999a) (there are 999 factors)
= 1/(2^999) * sin(2a)*sin(4a)*...
*sin(998a)*sin(1000a)*sin(1002a)*....*sin(1996a)*sin(1998a)
as sin(x)=-sin(2pi-x) and 2pi=1999a
sin(1000a)=-sin(2pi-1000a)=-sin(1999a-1000a)=-sin(999a)
sin(1002a)=-sin(2pi-1002a)=-sin(1999a-1002a)=-sin(997a)
...
sin(1996a)=-sin(2pi-1996a)=-sin(1999a-1996a)=-sin(3a)
sin(1998a)=-sin(2pi-1998a)=-sin(1999a-1998a)=-sin(a)
So sin(2a)*sin(4a)*...
*sin(998a)*sin(1000a)*sin(1002a)*....*sin(1996a)*sin(1998a)
= sin(a)*sin(2a)*sin(3a)*....*sin(998)*sin(999) since there are 500 sign "-".
Thus
P*Q=1/2^999*Q or Q!=0 then
P=1/(2^999)
We have been given that fuel efficiency for a 2007 passenger car was 31.2 mi/gal and the same model of car, the fuel efficiency increased to 35.6 mi/gal in 2012. Also, the gas tank for this car holds 16 gallons of gas.
We need to write a function and graph a linear function that models the distance that each car can travel for a given amount of gas up to one tankful.
Let represent the functions as
and
where
and
represent the distances traveled by car in years 2007 and 2012 and x represents the number of gallons. Therefore, we can express the required functions as:

Domain of both these functions are [0,16] and ranges are [0,499.2] and [0,569.6] respectively for years 2007 and 2012.
The difference function will be:


Domain of this function is [0,16] and range is [0,67.2].
The graphs are shown below.