First we find k by using the two values given:
P=Ae^(kt), 8 years between-->t=8
199 = 195•e^(8k)
Log 199 = Log [195•e^(8k)]
Log 199 = Log 195 + 8k
8k = Log 199 - Log 195
k = 0.00882/8 = .0011
Next, we plug the new data in using this k:
14 years between 2002-2016-->t=14
P = Ae^my
P = 199mi.•e^(14•.0011)
P = 199mi.•e^(.0154)
P = 199mi. • 1.0155 = 202,088,000
C+p=15
5c+7p=91
multiply 5 to 1st equation and subtract it from 2nd
multiply 5 to 1st equation to get
5c+5p=75
now subtract this from 2nd equation
thats 2p=16
so
p=8
now we find c from
c+p=15
c=7
16x
Step-by-step explanation:
6 increased by 2 --> 6+2
difference of 4 and 2 --> 4-2
first part times the second part --> 6 + 2 (4-2)
times a number --> • x
put it all together --> (6+2)(4-2) • x or 16x
What would you do if you ever needed to cook a chicken
if y is the time required to cook the bird and x is number of chicken lbs
y = 20x when x ≤ 6
y = 15x when x > 6
To Graph:
Have the y-axis (minutes) go by 5's since slopes of both equations are multiples of 5. The y-axis, I would suggest go to atleast 150 minutes for a 10 lbs chicken. The x-axis is in lbs and can go by 1 lb marks.
Start with the first equation where the chicken is smaller than or equal to 6 lbs. (0,0) (1,20), (2,40) (3, 60), (4,80), (5,100) to 6lbs (6,120). If the equation is EQUAL TO the point and the line will be SOLID.
for the second equation, the chicken is larger than 6 lbs. Start with x = 6 but the point will be OPEN and the line DASHED because it's NOT EQUAL TO. (6,90) (7,105) (8,120) (9,135), (10, 150) .... and on arrow to show this
20% of 124.95 is close to $125 so that is equal of 12.50 of 10% so multiply by 2 which is $25.00 so the estimate discount of the jacket will be $25.00
