If George completed 3/5 of work in 9 days, it means that he needs 3 days to finish 1/5th of the work.
The remaining part is 2/5 because 1-3/5=2/5. 2/5 is also 6/15
From this, how much did George do? in the first 3 days he did 1/5th and then one more day was left, during which he did 1/5/3=1/15th.
So he did in total 1/15+1/5=1/15+3/15=4/15.
this means that Paul did the rest of 6/15, so Paul did 6/15-4/15=2/15.
So we know that he does 2/15 in 4 days, which means that every two days he can do 1/15 of the work
so he would need 15 times 2 days to finish the work - 30 days!
52s + 30b because 52 x s + 30 x b = 52s + 30b
You could start by eliminating wrong answer(s):
(a) 5 is definitely not a factor of the given <span>2xy + 8y − 8x − 32,
(d) x^2 - 4x + 16 (not x2 - 4x + 16) is not a factor (it has no y terms)
On the other hand:
(b) 2 definitely divides evenly into </span>2xy + 8y − 8x − 32: 2(<span>xy + 4y − 4x − 16)
(c) x + 4 divides evenly into both xy + 4y and -(4x + 16):
We get 2(x +4) [y - 4]</span>
The nth taylor polynomial for the given function is
P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴
Given:
f(x) = ln(x)
n = 4
c = 3
nth Taylor polynomial for the function, centered at c
The Taylor series for f(x) = ln x centered at 5 is:

Since, c = 5 so,

Now
f(5) = ln 5
f'(x) = 1/x ⇒ f'(5) = 1/5
f''(x) = -1/x² ⇒ f''(5) = -1/5² = -1/25
f'''(x) = 2/x³ ⇒ f'''(5) = 2/5³ = 2/125
f''''(x) = -6/x⁴ ⇒ f (5) = -6/5⁴ = -6/625
So Taylor polynomial for n = 4 is:
P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴
Hence,
The nth taylor polynomial for the given function is
P₄(x) = ln5 + 1/5 (x-5) - 1/25*2! (x-5)² + 2/125*3! (x-5)³ - 6/625*4! (x - 5)⁴
Find out more information about nth taylor polynomial here
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Answer:
Step-by-step explanation:
Yes: 24 points/4 games = 6 points per game while 48 points/10 games = 24 points in 5 games.