It should be eight if I remember correctly
Might have to experiment a bit to choose the right answer.
In A, the first term is 456 and the common difference is 10. Each time we have a new term, the next one is the same except that 10 is added.
Suppose n were 1000. Then we'd have 456 + (1000)(10) = 10456
In B, the first term is 5 and the common ratio is 3. From 5 we get 15 by mult. 5 by 3. Similarly, from 135 we get 405 by mult. 135 by 3. This is a geom. series with first term 5 and common ratio 3. a_n = a_0*(3)^(n-1).
So if n were to reach 1000, the 1000th term would be 5*3^999, which is a very large number, certainly more than the 10456 you'd reach in A, above.
Can you now examine C and D in the same manner, and then choose the greatest final value? Safe to continue using n = 1000.
Answer:
StartRoot StartFraction 1 Over 41 EndFraction EndRoot
Negative StartRoot 41 EndRoot
Negative StartRoot StartFraction 1 Over 41 EndFraction EndRoot
StartRoot 41 EndRoot
The answer is b becuase my sister had that homework
<h3>Given :-</h3>
<h3>To find:</h3>
<h3>Solution:-</h3>
Let say it is first equation:-
x=y-1. . . . (1)
and this is second equation:-
x+2y=8 . . . . (2)
Simplifying 1 equation:-
- x = y - 1
- x +1 = y
- y = x + 1
Put this value of y in second equation.
to find value of y :-
y = x + 1
y = 2 + 1
y = 3
verification:-
1 equation:-
x = y - 1
put value of x and y
2 = 3 - 1
2 = 2
LHS = RHS
Hence verified!
2 equation:-
put value of x and y
- 2 + 2 × 3 = 8
- 2 + 6 = 8
- 8 = 8
LHS = RHS
Hence verified!
Both equation verified!
.°. value of x and y is 2 and 3 respectively
