1B: -3(9x-q)
-27x + 3q
Step 1: Distribute the -3 to the 9x to get -27x.
Step 2: Distribute the -3 to the -q.
2B: 2(12+5p)
24 + 10p
Step 1: Distribute the 2 to the 12 to get 24.
Step 2: Distribute the 2 to the 5p to get 10p.
3B: -7(-8y+7)
56y - 49
Step 1: Distribute the -7 to the -8y to get 56y.
Step 2: Distribute the -7 to the 7 to get -49.
4B: 10(2+7c)
20 + 70c
Step 1: Distribute the 10 to the 2 to get 20.
Step 2: Distribute the 10 to the 7c to get 70c.
5B: -8(7+11k)
-56 - 88k
Step 1: Distribute the -8 to the 7 to get -56.
Step 2: Distribute the -8 to the 11k to get -88k.
6B: -(3-7u)
-1(3-7u)
-3 + 7u
Step 1: Place a 1 after the negative symbol to symbolize -1.
Step 2: Distribute the -1 to the three to get -3.
Step 3: Distribute the -1 to the -7u to get 7u.
7B: -6(5p + s)
-30p - 6s
Step 1: Distribute the -6 to the 5p to get -30p.
Step 2: Distribute the -6 to the s to get -6s.
:) :D
Answer:
The median is 2
Step-by-step explanation:
Here, we want to get the median for the number of pencils
Firstly, we have to write out the numbers that form the box plot ; we have this as;
1,1,1,2,2,2,2,3,3,5
the median is sum of the 5th and 6th term divided by 2
The 5th term is 2
The 6th term is 2
So the median is;
(2 + 2)/2
= 4/2 = 2
B1 = 2
b2 = (b1)^2 + 1 = 2^2 + 1 = 5
b3 = (b2)^2 + 1 = 5^2 + 1 = 26
b4 = (b3)^2 + 1 = 26^2 + 1 = 676+1=<span>677</span>
Answer:
5, 6 and 7.
Step-by-step explanation:
5/8 is proper while 5/4 is improper
6/8 is proper while 6/4 is improper
7/8 is proper while 7/4 is improper
Answer:
The value of Car B will become greater than the value of car A during the fifth year.
Step-by-step explanation:
Note: See the attached excel file for calculation of beginning and ending values of Cars A and B.
In the attached excel file, the following are used:
Annual Depreciation expense of Car A = Initial value of Car A * Depreciates rate of Car A = 30,000 * 20% = 6,000
Annual Depreciation expense of Car B from Year 1 to Year 6 = Initial value of Car B * Depreciates rate of Car B = 20,000 * 15% = 3,000
Annual Depreciation expense of Car B in Year 7 = Beginning value of Car B in Year 7 = 2,000
Conclusion
Since the 8,000 Beginning value of Car B in Year 5 is greater than the 6,000 Beginning value of Car A in Year 5, it therefore implies that the value Car B becomes greater than the value of car A during the fifth year.
