Answer:
Thx a lot
Step-by-step explanation:
Answer: Yes
Step-by-step explanation:
2x+8y=6
-8y -8y
2x = -2y
Divide by two so x will be by itself, so the answer will be -1y.
5x+20y=2
-20y -20y
5x = -18y
Divide by 5 so x will be by itself, so the answer will be -3y.
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
We can adjust the data by adding 4 to everything before we calculate the statistics. Or we can calculate the statistics on the given data and just add 4 to everything at the end. We'll get the same answer either way.
Let's sort the seven data points: 5 5 5 7 7 9 10
Those add up to 48 so the mean is 48/7 = 6.9
The one in the middle is 7 so the median = 7
The mode is the most common one, mode = 5
The range is the difference between max and min, so range = 10 - 5 = 5
In the second week we add four to everything. Since that adds four to the min and max, the range doesn't change.
Answer: mean=10.9, median=11, mode=9, range=5
Answer:
3 and 4
1 and 3
Step-by-step explanation:
This is because these are vertical angles so anything on one side is the same on the other in these words Congruent. Good luck.
