Answer:

So then P =11000 is the minimum that the least populated district could have.
Step-by-step explanation:
We have a big total of N = 132000 for the population.
And we know that we divide this population into 11 districts
And we have this info given "no district is to have a population that is more than 10 percent greater than the population of any other district"
Let's assume that P represent our minimum value for a district in the population. The range of possible values for the population of each district would be between P and 1.1 P
The interest on this case is find the minimum value for P and in order to do this we can assume that 1 district present the minimum and the other 10 the maximum value 1.1P in order to find which value of P satisfy this condition, and we have this:


So then P =11000 is the minimum that the least populated district could have.
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
You do three fourths times eight and that equals six.
Brianna's thinking is wrong because obviously all of the expressions are going to equal -4 when x is 0 because -4 would be the only value. Also, if x was a different number, the expressions wouldn't be equivalent. The equivalent expressions are A. 9x - 3x - 4, and C. 5x + x - 4. This is because when both are simplified, they equal 6x - 4.