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Ksenya-84 [330]
2 years ago
8

In what year did the nfl vote to create a $430 million pool of money, funded by the top 15 clubs by revenue to be distributed to

the bottom 15 clubs?.
Mathematics
1 answer:
kotykmax [81]2 years ago
3 0

The year in which the NFL voted to create a $430 million fund to help the bottom 15 clubs was <u>2007.</u>

<h3>When did the NFL vote to implement this fund?</h3>

The NFL voted by a margin of 30-2 in 2007 to create a pool of funds over 3 years that would amount to $430 million.

This money would then be used to help the bottom 15 clubs who might be struggling financially. The hope was that this would make things more competitive in the NFL.

Find out more on the NFL at brainly.com/question/7579515.

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it takes 1/3 cup of sugar to make a milkshake you have 6 cups of sugar how many milk shakes can you make?​
Gekata [30.6K]

Answer:

18

Step-by-step explanation:

1/3 * 18 = 6 or 6/(1/3) = 18 or 6 * 3 =18

6 0
3 years ago
When simplifying an expression using order of operations addition MUST
Rama09 [41]

Answer:

The answer is False,

Step-by-step explanation:

The answer is False because in some expressions if there are parentheses and there is a subtraction problem in the parentheses but there is an addition problem in front of the parentheses that does not exactly mean that you do the addition first, this is because the subtraction is inside the parentheses and so since the subtraction is in parentheses it is done fist.

4 0
2 years ago
Consider the function f(x)=xln(x). Let Tn be the nth degree Taylor approximation of f(2) about x=1. Find: T1, T2, T3. find |R3|
Fynjy0 [20]

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

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

From jamsie pooh

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