Answer:
combine like terms.
Step-by-step explanation:
none of the other answers seem correct.
Answer:
6
Step-by-step explanation:
If x is the number of 1/8 yard long pieces, then the total length is x * 1/8. We know that the total length is 3/4 yards. Therefore:
x * 1/8 = 3/4
Divide:
x = (3/4) / (1/8)
To divide a fraction, multiply by its reciprocal:
x = (3/4) * (8/1)
x = 24/4
x = 6
She has 6 pieces.
Answer:
Step-by-step explanation:
![\ln \dfrac{4y^5}{x^2}\\\\=\ln(4y^5) - \ln(x^2)~~~~~~~~~~~;\left[ \log_b\left( \dfrac mn \right) = \log_b m - \llog_b n \right]\\\\=\ln 4 + \ln y^5 - 2\ln x~~~~~~~~~~~~;[\log_b m^n = n \log_b m ~\text{and}~\log_b(mn) = \log_b m + \log_b n ]\\\\=\ln 4 + 5 \ln y -2 \ln x\\\\=\ln 4 -2 \ln x +5 \ln y](https://tex.z-dn.net/?f=%5Cln%20%5Cdfrac%7B4y%5E5%7D%7Bx%5E2%7D%5C%5C%5C%5C%3D%5Cln%284y%5E5%29%20-%20%5Cln%28x%5E2%29~~~~~~~~~~~%3B%5Cleft%5B%20%5Clog_b%5Cleft%28%20%5Cdfrac%20mn%20%5Cright%29%20%20%3D%20%5Clog_b%20m%20-%20%5Cllog_b%20n%20%5Cright%5D%5C%5C%5C%5C%3D%5Cln%204%20%2B%20%5Cln%20y%5E5%20-%202%5Cln%20x~~~~~~~~~~~~%3B%5B%5Clog_b%20m%5En%20%3D%20n%20%5Clog_b%20m%20~%5Ctext%7Band%7D~%5Clog_b%28mn%29%20%3D%20%5Clog_b%20m%20%2B%20%5Clog_b%20n%20%5D%5C%5C%5C%5C%3D%5Cln%204%20%2B%205%20%5Cln%20y%20-2%20%5Cln%20x%5C%5C%5C%5C%3D%5Cln%204%20-2%20%5Cln%20x%20%2B5%20%5Cln%20y)
Answer:
274560
Step-by-step explanation:
We can "choose" one of the 66 athletes for the gold medal. That athlete can't win any other medals, so there are 65 athletes left.
We can then "choose" one of the 65 athletes left for the silver medal. That athlete can't win any other medals, so there are 64 athletes left.
We can then "choose" one of the 64 athletes left for the bronze medal.
That leaves us with 66 possible choices * 65 possible choices * 64 possible choices=274560 possible choices
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
