The y value of I is (5(2) + 3(5))/(3 + 5) = (10 + 15)/8 = 25/8 = 3.125
these are just two semi-circles, one with a diameter of 22 and the other with a diameter of 30 - 22, or 8, and therefore their radius are 11 and 4 respectively.
Cos35= y/15
y= 15cos35
y= 12.3 cm
The diagonal of a rectangle = sqrt(w^2 + l^2)
w = width
l = length
In this problem,
The diagonal = 20 in
w = x
l = 2x + 8
Let's plug our numbers into the formula above.
20in = sqrt((x)^2 + (2x + 8)^2)
Let's simplify the inside of the sqrt
20 in = sqrt(5x^2 + 32x + 64)
Now, let's square both sides.
400 = 5x^2 + 32x + 64
Subtract 400 from both sides.
0 = 5x^2 + 32x - 336
Factor
0 = (5x - 28)(x + 12)
Set both terms equal to zero and solve.
x + 12 = 0
Subtract 12 from both sides.
x = -12
5x - 28 = 0
Add 28 to both sides.
5x = 28
Divide both sides by 5
x = 28/5
The width cant be a negative number so now we know that the only real solution is 28/5
Let's plug 28/5 into our length equation.
Length = 2(28/5) + 8 = 56/5 + 8 = 96/5
In conclusion,
Length = 96/5 inches
Width = 28/5
1/7 = 2/14 =
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0.1428571429
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1/8 = 2/16 = 0.125
2/15 =
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0.1333333333
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So, 2/15 is between 1/7 and 1/8
(and there are an infinite amount of numbers between 1/7 and 1/8)
