Answer:
First term (a) =8
Common difference (d)= t2-t1
=12-8
=4
Now, sum of first 31th term (tn31) =n/2{2a+(n-1)d}
= 31/2{2×8+(31-1)4}
=31/2{16+(30×4)
=31/2(16+120)
=31/2×126
=31×63
Step-by-step explanation:
Similarly use 19 as (n) for the 19th term
The answer for this problem i −10x2+66xy−36y2−5x+3y
Let's solve it by step by step.
(−5x+3y)(2x−12y+1)
=(−5x+3y)(2x+−12y+1)
=(−5x)(2x)+(−5x)(−12y)+(−5x)(1)+(3y)(2x)+(3y)(−12y)+(3y)(1)
=−10x2+60xy−5x+6xy−36y2+3y
So, -10x2+66xy-36y2-5x+3y
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Answer:
No
Step-by-step explanation:
Neither 4 or 7 is a solution to both equations
Answer:
20qrt/s / as in over
Step-by-step explanation:
9514 1404 393
Answer:
(x, y) = (7, -48)
Step-by-step explanation:
A critical point is where the slope is zero or undefined. A polynomial function will never have any points where the slope is undefined.
The slope is zero were f'(x) = 0.
f'(x) = 2x -14
0 = 2x -14
0 = x -7
7 = x
f(7) = 49 -98 +1 = -48
The critical point is (7, -48).