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TiliK225 [7]
3 years ago
13

How many weeks will it take to play 20 tournaments

Mathematics
1 answer:
zimovet [89]3 years ago
8 0
I think it would depend on how many tournaments you play a week
You might be interested in
What two consecutive numbers have a sum of 53? Thanks! :P
Musya8 [376]
Let's the name the first number x and the consecutive number x + 1. The sum of both of these numbers equals to 53.

We now have our equation:

x + x + 1 = 53

Now solve for x.

x + x + 1 = 53
2x + 1 = 53 <-- Combine like terms
2x = 52 <-- Subtract 1 from each side
x = 26

So, the first number is 26 and the second number is 27.
4 0
4 years ago
Read 2 more answers
Find the slope and y-intercept for 2x+8y=-32 and then graph the line
sammy [17]
2x+8y=-32
8y=-2x-32
y=-1/4x-4

The slope is -1/4, the y-intercept is -4, and I can't graph it for you on here. Type my equation into a graphing calculator and it'll do it for you. (Desmos is a good online graphing calc, and it's free)
8 0
4 years ago
Express 6.925 x 10^–5 in standard form.
KiRa [710]

es la de por qué si no es nada malooo y tu qué haces con tu

5 0
3 years ago
Read 2 more answers
A swimming pool is 20 feet wide and 40 feet long. If it is surrounded by a walkway of square tiles, each is 1 foot by 1 foot, ho
Lena [83]

Answer:

124 tiles

Step-by-step explanation:

2(20+40)

2(60)

120+4=124

5 0
3 years ago
How do you do this question?
lukranit [14]

Answer:

0.001591

Step-by-step explanation:

The power series for arctan(x) is:

arctan(x) = ∑ (-1)ⁿ x²ⁿ⁺¹ / (2n + 1)

Substituting 5x:

arctan(5x) = ∑ (-1)ⁿ (5x)²ⁿ⁺¹ / (2n + 1)

Multiply both sides by x:

x arctan(5x) = ∑ (-1)ⁿ x (5x)²ⁿ⁺¹ / (2n + 1)

Simplify:

x arctan(5x) = ∑ (-1)ⁿ (5x) (5x)²ⁿ⁺¹ / (10n + 5)

x arctan(5x) = ∑ (-1)ⁿ (5x)²ⁿ⁺² / (10n + 5)

Multiply top and bottom by 5:

x arctan(5x) = ∑ (-1)ⁿ 5 (5x)²ⁿ⁺² / (50n + 25)

Integrate:

∫ x arctan(5x) = ∑ (-1)ⁿ (5x)²ⁿ⁺³ / ((50n + 25) (2n + 3))

Evaluate between x = 0.1 and x = 0:

∫₀⁰¹ x arctan(5x) = [∑ (-1)ⁿ (5x)²ⁿ⁺³ / ((50n + 25) (2n + 3))] |₀⁰¹

∫₀⁰¹ x arctan(5x) = ∑ (-1)ⁿ (0.5)²ⁿ⁺³ / ((50n + 25) (2n + 3))

This is an alternating series.  We can approximate it with Alternating Series Estimation.

bₙ₊₁ ≥ ε

(0.5)²⁽ⁿ⁺¹⁾⁺³ / ((50(n+1) + 25) (2(n+1) + 3)) ≥ 0.000001

(0.5)²ⁿ⁺⁵ / ((50n + 75) (2n + 5)) ≥ 0.000001

n ≥ 3

So the approximation is the sum of the terms from n=0 to n=3.

(-1)⁰ (0.5)²⁽⁰⁾⁺³ / ((50(0) + 25) (2(0) + 3))

+ (-1)¹ (0.5)²⁽¹⁾⁺³ / ((50(1) + 25) (2(1) + 3))

+ (-1)² (0.5)²⁽²⁾⁺³ / ((50(2) + 25) (2(2) + 3))

+ (-1)³ (0.5)²⁽³⁾⁺³ / ((50(3) + 25) (2(3) + 3))

= 0.0016667 − 0.0000833 + 0.0000089 − 0.0000012

= 0.001591

7 0
3 years ago
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