Answer:
see explanation
Step-by-step explanation:
The common difference d of an arithmetic sequence is
d = 
- 
= 
-
Substitute in values and solve for k, that is
5k - 1 - 2k = 6k + 2 - (5k - 1)
3k - 1 = 6k + 2 - 5k + 1
3k - 1 = k + 3 ( subtract k from both sides )
2k - 1 = 3 ( add 1 to both sides )
2k = 4 ⇒ k = 2
--------------------------------------------------------
The n th term of an arithmetic sequence is
= + (n - 1)d
= 2k = 2 × 2 = 4 and
d = 5k - 1 - 2k = 3k - 1 = (3 × 2) - 1 = 5
Hence
= 4 + (7 × 5) = 4 + 35 = 39
I believe this is how you do it
Answer:
Increasing if f' >0 and decreasing if f'<0
Step-by-step explanation:
Difference quotient got by getting
will be greater than 0 if function is increasing otherwise negative
Here h is a small positive value.
In other words, we find that whenever first derivative of a function f(x) is positive the function is increasing.
Here given that for x1, x2 where x1<x2, we have
if f(x1) <f(x2) then the function is decreasing.
Or if x1<x2 and if f(x1) >f(x2) for all x1, and x2 in I the open interval we say f(x) is decreasing in I.
Answer:
y=-3x+1
Step-by-step explanation:
plug in point into x and y, replace -4 as a +b(or any other variable to represent y intercept)
therefore,
7=-3(-2)+b
7=6+b
1=b
y=-3x+1
Answer:
See the three attached images.
Step-by-step explanation:
image1 shows the intersection of b and c: 
. (green "football")
image2 shows the <u>complement</u> of a (outside a): 
(yellow)
image3 shows the intersection of those sets: 
(green "football with a bite out of it") :-)
