Answer:
Part (A): The total number of ways are
Part (B): The total number of ways are
Step-by-step explanation:
Consider the provided information.
Part(A) a) How many ways are there for her to plan her schedule if there are no restrictions on the number of days she studies each of the four subjects?
She plans on studying four different subjects.
That means she has 4 choices for each day also there is no restriction on the number of days.
Hence, the total number of ways are:
Part (B) How many ways are there for her to plan her schedule if she decides that the number of days she studies each subject will be the same?
She wants that the number of day she studies each subject will be the same that means she studies for 25 days each subject.
Therefore, the number of ways are:
Answer:
x = 14
Step-by-step explanation:
∠ NQM and ∠ PQR are congruent ( indicated by the same orange arc within the 2 angles ) , then
2x + 8 = x + 22 ( subtract x from both sides )
x + 8 = 22 ( subtract 8 from both sides )
x = 14
Step-by-step explanation:
Given (2x + 23), (8x + 2) and (20x - 52) are three consecutive terms of an arithmetic sequence.
(8x + 2) - (2x + 23) = (20x - 52) - (8x + 2)
or, 6x - 21 = 12x - 54
or, 12x - 6x = - 21 + 54 = 33
or, 6x = 33
or, 2x = 11
∴ x = 11/2
∴2x + 23 = 2 × 11/2 + 23 = 34
8x + 2 = 8 × 11/2 + 2 = 46 and
20x - 52 = 20 × 11/2 - 52 = 110 - 52 = 58
34, 46 and 58 are three consecutive terms of an arithmetic sequence.
∴ Common difference(d) = 46 - 34 = 58 - 46 = 12, it is proved.
Hence, the common difference of the sequence is 12.
<span>I am almost down with advanced functions and I found it to be a little difficult. I've never been that strong in math so I expected this course to be hard for me. I tried pretty hard this year, and I think after the exam, I can maintain around a 85-87. I'm just wondering if calculus is a lot more difficult than functions to the people that have already taken both courses. I've heard multiple times that calculus can be easier, but I'm not quite sure.</span>