Answer:
y
=
x
(2)
+
10
x
+
24
Step-by-step explanation:
quadrent 3
9514 1404 393
Answer:
5) 729, an=3^n, a[1]=3; a[n]=3·a[n-1]
6) 1792, an=7(4^(n-1)), a[1]=7; a[n]=4·a[n-1]
Step-by-step explanation:
The next term of a geometric sequence is the last term multiplied by the common ratio. (This is the basis of the recursive formula.)
The Explicit Rule is ...
for first term a₁ and common ratio r.
The Recursive Rule is ...
a[1] = a₁
a[n] = r·a[n-1]
__
5. First term is a₁ = 3; common ratio is r = 9/3 = 3.
Next term: 243×3 = 729
Explicit rule: an = 3·3^(n-1) = 3^n
Recursive rule: a[1] = 3; a[n] = 3·a[n-1]
__
6. First term is a₁ = 7; common ratio is r = 28/7 = 4.
Next term: 448×4 = 1792
Explicit rule: an = 7·4^(n-1)
Recursive rule: a[1] = 7; a[n] = 4·a[n-1]
Answer:
R3 <= 0.083
Step-by-step explanation:
f(x)=xlnx,
The derivatives are as follows:
f'(x)=1+lnx,
f"(x)=1/x,
f"'(x)=-1/x²
f^(4)(x)=2/x³
Simialrly;
f(1) = 0,
f'(1) = 1,
f"(1) = 1,
f"'(1) = -1,
f^(4)(1) = 2
As such;
T1 = f(1) + f'(1)(x-1)
T1 = 0+1(x-1)
T1 = x - 1
T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2
T2 = 0+1(x-1)+1(x-1)^2
T2 = x-1+(x²-2x+1)/2
T2 = x²/2 - 1/2
T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3
T3 = 0+1(x-1)+1/2(x-1)^2-1/6(x-1)^3
T3 = 1/6 (-x^3 + 6 x^2 - 3 x - 2)
Thus, T1(2) = 2 - 1
T1(2) = 1
T2 (2) = 2²/2 - 1/2
T2 (2) = 3/2
T2 (2) = 1.5
T3(2) = 1/6 (-2^3 + 6 *2^2 - 3 *2 - 2)
T3(2) = 4/3
T3(2) = 1.333
Since;
f(2) = 2 × ln(2)
f(2) = 2×0.693147 =
f(2) = 1.386294
Since;
f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).
Then; we have, R3 <= | f^(4)(c)/(4!)(x-1)^4 |,
Since;
f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2
Finally;
R3 <= |2/(4!)(2-1)^4|
R3 <= | 2 / 24× 1 |
R3 <= 1/12
R3 <= 0.083
The answer is C,
I think that’s eighth
Answer:
A
Y-Intercept is -2
X then Y so it would be -2,-4
