Answer:
Σ(-1)^kx^k for k = 0 to n
Step-by-step explanation:
The nth Maclaurin polynomials for f to be
Pn(x) = f(0) + f'(0)x + f''(0)x²/2! + f"'(0)x³/3! +. ......
The given function is.
f(x) = 1/(1+x)
Differentiate four times with respect to x
f(x) = 1/(1+x)
f'(x) = -1/(1+x)²
f''(x) = 2/(1+x)³
f'''(x) = -6/(1+x)⁴
f''''(x) = 24/(1+x)^5
To calculate with a coefficient of 1
f(0) = 1
f'(0) = -1
f''(0) = 2
f'''(0) = -6
f''''(0) = 24
Findinf Pn(x) for n = 0 to 4.
Po(x) = 1
P1(x) = 1 - x
P2(x) = 1 - x + x²
P3(x) = 1 - x+ x² - x³
P4(x) = 1 - x+ x² - x³+ x⁴
Hence, the nth Maclaurin polynomials is
1 - x+ x² - x³+ x⁴ +.......+(-1)^nx^n
= Σ(-1)^kx^k for k = 0 to n
Answer: We have two solutions:
1000 - 998 = 2
1001 - 999 = 2
Step-by-step explanation:
So we have the problem:
****-*** = 2
where each star is a different digit, so in this case, we have a 4 digit number minus a 3 digit number, and the difference is 2.
we know that if we have a number like 99*, we can add a number between 1 and 9 and we will have a 4-digit as a result:
So we could write this as:
1000 - 998 = 2
now, if we add one to each number, the difference will be the same, and the number of digits in each number will remain equal:
1000 - 998 + 1 - 1 = 2
(1000 + 1) - (998 + 1) = 2
1001 - 999 = 2
now, there is a trivial case where we can find other solutions where the digits can be zero, like:
0004 - 0002 = 2
But this is trivial, so we can ignore this case.
Then we have two different solutions.
Answer:
Option A.
Step-by-step explanation:
The given expression is
![\dfrac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B7%5D%7Bx%5E2%7D%7D%7B%5Csqrt%5B5%5D%7By%5E3%7D%7D)
where, 
.
We need to find the expression which is equivalent to the given expression.
The given expression can be rewritten as
![[\because \sqrt[n]{x}=x^{\frac{1}{n}}]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Csqrt%5Bn%5D%7Bx%7D%3Dx%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D%5D)
![[\because (a^m)^n=a^{mn}]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%28a%5Em%29%5En%3Da%5E%7Bmn%7D%5D)
![[\because a^{-n}=\dfrac{1}{a^n}]](https://tex.z-dn.net/?f=%5B%5Cbecause%20a%5E%7B-n%7D%3D%5Cdfrac%7B1%7D%7Ba%5En%7D%5D)
Therefore, the correct option is A.
<h2>
Answer:</h2>
Option: B is the correct answer.
The range of the function is:
B. 5 < y < ∞
<h2>
Step-by-step explanation:</h2>
Range of a function--
The range of a function is the set of all the values that is attained by the function.
By looking at the graph of the function we see that the function tends to 5 when x→ -∞ and the function tends to infinity when x →∞
Also, the function is a strictly increasing function.
This means that the function takes every real value between 5 and ∞ .
i.e. The range of the function is: (5,∞)
Hence, the answer is:
Option: B
To solve this problem, you simply have to do base 1 * height 1 = base 2 * height 2. Plugging our numbers in gives us 24 * 4 = 5 * height 2. Solving it out, we get 19.2 as our missing height.
