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Margarita [4]
3 years ago
10

Convert the fractions to decimals.

Mathematics
1 answer:
AnnZ [28]3 years ago
5 0
<span>3/20: 0.15
7/50: 0.14
9/25: 0.36
4/15: 0.266667
1/9: 0.111111
9/40: 0.225
5/16: 0.3125
7/9: 0.777778
13/20: 0.65
37/50: 0.74
11/30: 0.366667
19/40: </span>0.475
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8 9 10 11 12 13 14 15 16 17 18 19 20You and two friends (Adam and Alana) wonly play a game if it is fair for all three of you. T
pav-90 [236]
Well I don't know ! 
Let's take a look and see:

The idea is that there could be more than one way
for a roll of the dice to land with the same number.  

-- If the sum is from 1-4, you get the point.
There are  6  different ways for a roll of the dice to come up 1-4.

-- If the sum is from 5-8, Adam gets the point.
There are  20  different ways for a roll of the dice to come up 5-8.

-- If the sum is 9-12, Lana gets the point.
There are  10  different ways for a roll of the dice to come up 9-12.

-- The game is not fair to all three of you.
-- Lana has a distinct advantage over you.
-- Adam has a big advantage over Lana.
-- Adam has an even bigger advantage over you.
-- You are at a big disadvantage.  (Notice that one of your
numbers ... 1 ... can never come up unless one of the dice
falls off of the table.)
_______________________________

Here's how to figure it:

Ways to roll a 2:
1 ... 1

Ways to roll a 3:
1 ... 2
2 ... 1

Ways to roll a 4:
1 ... 3
2 ... 2
3 ... 1

Ways to roll a 5:
1 ... 4
2 ... 3
3 ... 2
4 ... 1

Ways to roll a 6:
1 ... 5
2 ... 4
3 ... 3
4 ... 2
5 ... 1

Ways to roll a 7:
1 ... 6
2 ... 5
3 ... 4
4 ... 3
5 ... 2
6 ... 1

Ways to roll an 8:
2 ... 6
3 ... 5
4 ... 4
5 ... 3
6 ... 2

Ways to roll a 9:
3 ... 6
4 ... 5
5 ... 4
6 ... 3

Ways to roll a 10:
4 ... 6
5 ... 5
6 ... 4

Ways to roll 11:
5 ... 6
6 ... 5

Ways to roll 12:
6 ... 6

5 0
3 years ago
Boxes that are 12 inches tall are being stacked next toboxes that are 18 inches tall. What is the shortest height in which the t
ycow [4]
36 inches because 12,24,36 and 18,36, 36=36
6 0
3 years ago
One of the vertices of an equilateral triangle is on the vertex of a square and two other vertices are on the not adjacent sides
Elina [12.6K]
<h2>Answer:</h2>

<em> The side of the triangle is either 38.63ft or 10.35ft</em>

<h2>Step-by-step explanation:</h2>

This problem can be translated as an image as shown in the Figure below. We know that:

  • The side of the square is 10 ft.
  • One of the vertices of an equilateral triangle is on the vertex of a square.
  • Two other vertices are on the not adjacent sides of the same square.

Let's call:

Since the given triangle is equilateral, each side measures the same length. So:

x: The side of the equilateral triangle (Triangle 1)

y: A side of another triangle called Triangle 2.

That length is the hypotenuse of other triangle called Triangle 2. Therefore, by Pythagorean theorem:

\mathbf{(1)} \ x^2=100+y^2

We have another triangle, called Triangle 3, and given that the side of the square is 10ft, then it is true that:

y+(10-y)=10

Therefore, for Triangle 3, we have that by Pythagorean theorem:

(10-y)^2+(10-y)^2=x^2 \\ \\ 2(10-y)^2=x^2 \\ \\ \\ \mathbf{(2)} \ x^2=2(10-y)^2

Matching equations (1) and (2):

2(10-y)^2=100+y^2 \\ \\ 2(100-20y+y^2)=100+y^2 \\ \\ 200-40y+2y^2=100+y^2 \\ \\ (2y^2-y^2)-40y+(200-100)=0 \\ \\ y^2-40y+100=0

Using quadratic formula:

y_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ y_{1,2}=\frac{-(-40) \pm \sqrt{(-40)^2-4(1)(100)}}{2(1)} \\ \\ \\ y_{1}=37.32 \\ \\ y_{2}=2.68

Finding x from (1):

x^2=100+y^2 \\ \\ x_{1}=\sqrt{100+37.32^2} \\ \\ x_{1}=38.63ft \\ \\ \\ x_{2}=\sqrt{100+2.68^2} \\ \\ x_{2}=10.35ft

<em>Finally, the side of the triangle is either 38.63ft or 10.35ft</em>

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