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
.
So,


Say if you wanted to find factors of 6, the you'd multiply that number by any number to find a factor or multiple.
Question 1:
73 is a prime number. It can only be divided by 1 and by itself.
The GCF of the three numbers:
54 36 73
1×54 1×36 1×73
2×27 2×18
3×18 3×12
6×9 4×9
6×6
GCF of 54, 36 and 73 is 1
GCF of 54 and 36 is 18
If we divide 54 apples into 18 baskets, we have 3 apples in each basket
If we divide 36 oranges into 18 baskets, we have 2 oranges in each basket
If we divide 73 bananas into 18 baskets, we have 4 bananas in each basket + one banana left over.
So the greatest number of identical fruit baskets we can make with the least amount of fruit left over is 18 baskets
Each bottle serves 2 students. He needs to buy 14/2 = 7 bottles.
Answer:
a. yes
b. no
c. yes
d. yes
Step-by-step explanation:
For a, 27 + 38 can be broken apart. 27 is broken up by adding smaller numbers (20 + 7 = 27) and the same is done with 38 (30 + 8 = 38), so A and C shows a way to add 27 and 38. In C, the numbers are just put into a different order. B is not a way to add 27 and 38, because the sum is different.
27 + 38 = 65, however 20 + 70 + 38 = 128. The addition problem for D is a way to solve for 27 + 38, because it is broken up differently than A and C. They instead added the 20 and 30 together first, then split up 15 (from 8+7) into 10 and 5. So D is a way to solve, because it gets the same answer as 27 and 38 :D