Answer:
2 11/15
Step-by-step explanation:
Let X equal the decimal number
Equation 1:
X=2.73¯¯¯(1)
With 1 digits in the repeating decimal group,
create a second equation by multiplying
both sides by 101 = 10
Equation 2:
10X=27.33¯¯¯(2)
Subtract equation (1) from equation (2)
10XX9X===27.33...2.73...24.6
We get
9X=24.6
Solve for X
X=24.69
Multiply to eliminate 1 decimal places.
Here you multiply top and bottom by 1 10's
= 101 = 10
24.69×1010=24690
Find the Greatest Common Factor (GCF) of 246 and 90, if it exists, and reduce the fraction by dividing both numerator and denominator by GCF = 6,
246÷690÷6=4115
Simplify the improper fraction,
=21115
In conclusion,
2.73¯¯¯=2 11/15
<h3>
Answer: 14x - 8</h3>
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Explanation:
I'll use the quadratic formula to find the roots or x intercepts. This slight detour allows us to factor without having to use guess-and-check methods.
The equation is of the form ax^2+bx+c = 0
This leads to...
Now use those roots to form these steps
Refer to the zero product property for more info.
Therefore, the original expression factors fully to (4x-5)(3x+1)
Use the FOIL rule to expand it out and you should get 12x^2-11x-5 again.
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We did that factoring so we could find the side lengths of the rectangle.
I'm using the fact that area = length*width
- L = length = 4x-5
- W = width = 3x+1
The order of length and width doesn't matter.
From here, we can then compute the perimeter of the rectangle
P = 2(L+W)
P = 2(4x-5+3x+1)
P = 2(7x-4)
P = 14x - 8
Answer:
y=ln(x/(1-x))
Step-by-step explanation:
y=e^x/(1+e^x)
Cross multiply
y(1+e^x)=e^x
Distribute
y+ye^x=e^x
Put anything with x on with side and everything without x on opposing side:
y=e^x-ye^x
Factor right hand side
y=(1-y)e^x
Divide both sides by (1-y)
y/(1-y)=e^x
Use natural log.
ln(y/(1-y))=x
The inverse is
y=ln(x/(1-x))
I would prefer to solve this in a decimal form
I will show this to u in 2 ways
1:.75::X:2.25
(1*2.25)/.75x
2.25/.75x
X=3
2.25/.75=3
when we divide the number of hours played by hors for one game we get the number of games that could be played
