1/5, 3/4, 83/100.
Hope this helps you...
Split up the interval [0, 2] into 4 subintervals, so that
![[0,2]=\left[0,\dfrac12\right]\cup\left[\dfrac12,1\right]\cup\left[1,\dfrac32\right]\cup\left[\dfrac32,2\right]](https://tex.z-dn.net/?f=%5B0%2C2%5D%3D%5Cleft%5B0%2C%5Cdfrac12%5Cright%5D%5Ccup%5Cleft%5B%5Cdfrac12%2C1%5Cright%5D%5Ccup%5Cleft%5B1%2C%5Cdfrac32%5Cright%5D%5Ccup%5Cleft%5B%5Cdfrac32%2C2%5Cright%5D)
Each subinterval has width 
. The area of the trapezoid constructed on each subinterval is 
, i.e. the average of the values of 
at both endpoints of the subinterval times 1/2 over each subinterval ![[x_i,x_{i+1}]](https://tex.z-dn.net/?f=%5Bx_i%2Cx_%7Bi%2B1%7D%5D)
.
So,
Answer:
The answer to your question is:
Packages of pencils = 6
Packages of erasers = 5
Step-by-step explanation:
Data
Pencils = 10/package
Erasers = 12 / package
Process
Find the least common factor of 10 and 12
10 12 2
5 6 2
5 3 3
5 1 5
1
LCF = 2 x 2 x 3 x 5 = 60
Finally divide 60 by the number of pencils or erasers in each package
Packages of pencils = 60/10 = 6
Packages of erasers = 60/12 = 5
Answer:1700
Step-by-step explanation:
The slope is negative and is -16 / 8 = -2
The y intercept is at y = 16
General form is y = mx + c
m = -2 and c = 16 so we have :-
y = -2x + 16 Answer
