The answer is B. Yes; 4
Arithmetic sequence is the type of sequence of numbers which difference between each number will always remain constant
in the sequence above we could see the difference between one number with another is exactly 4:
15 + 4 = 19
19 + 4 = 23
23 + 4 = 27
Answer:
In what article are you most likely to find the sentence “The coma streamed out until it formed a tail of dust and light”?
“Comets in the Sky”
“How to Punctuate Sentences”
“Water Under the Bridge”
“Medical Histories”
Step-by-step explanation:
Answer:
cups of butter should be use to keep the same ratio.
Step-by-step explanation:
Given:
A recipe for pie dough calls for a ratio of butter to flour of 1/4 :5/6.
If you plan to use 5 cups of flour.
Now, to find the butter should be used to keep the same ratio.
So, to get the quantity of butter by using unitary method:
If, 
cup of flour use then butter should be = 
.
So, 1 cup of flour use then butter should be = 
= 
.
Thus, for 5 cups of flour butter should be = 
= 
.
Therefore, cups of butter should be use to keep the same ratio.
X + y = 27
x = y + 3
y + 3 + y = 27
2y + 3 = 27
2y = 27 - 3
2y = 24
y = 24/2
y = 12
x = y + 3
x = 12 + 3
x = 15
ur numbers are 12 and 15
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.
