Pretty much, write two problems that equal to 0
If you graph the end points C and D then graph the 4 points at the end it is difficult to tell which points are on CD without a line.
Using the endpoints find the slope (change in y/ change in x) then substitute a point in to find the intercept.
Slope = (-6-4)/(6- -8) = -5/7
Intercept equation (-6) = -5/7 (6) + b
b = -1.71428571429
Graphing the line shows only 2 points on the line (–2.75, 0.25) and <span>(0.75, –2.25)
I am confused by the part, "</span><span>P is the length of the line segment from D". Were you given a length P to help you determine which point. Using the distance formula to find the length from each point to D doesn't help determine which one is best with the information you have given. The image shows the distances I calculated and the graphed points.
I hope this helps!</span>
The common difference of the sequence is -3 and the fifth term is 5
<h3>How to determine the common difference?</h3>
The sequence is given as:
17, 14, 11, 8....
The common difference is
d = T2 - T1
So, we have
d = 14 - 17
Evaluate
d = -3
Hence, the common difference of the sequence is -3
<h3>How to determine the
fifth term?</h3>
The fifth term is calculated as:
T5 = a + 4d
Where
a = T1 = 17
d = -3
So, we have:
T5 = 17 - 4 * 3
Evaluate
T5 = 5
Hence, the fifth term is 5
<h3>How to determine the
nth term?</h3>
The nth term is calculated as:
Tn = a + (n - 1)d
Hence, the nth term is Tn = a + (n - 1)d
Read more about arithmetic sequence at:
brainly.com/question/6561461
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Answer:
Step-by-step explanation:
a) We have 15! as the product of 1 to 15 natural numbers. Since 17 is prime there will be no factor common to these
By actual division we find
15! (mod 17) =16
From this we deduce
even 16! mod 17 = 16 = -1
According to Wilson theorem
(17-1)! = -1 mod 17
Thus verified 17 is prime
Hence 15! (mod 17) =-1=16
-----------------------
b) 2(26!) is divided by 29
Since 29 is prime
(29-1)! = -1 mod 29
28! = -1 mod 29 = 28
When divided this gives 25 as remainder
I assume you mean
, and not
, since this doesn't satisfy the ODE.
Assume a second solution of the form
, where
is a function of
. Then


Substituting into the ODE gives







where we omit the second term because it's already accounted for by
.