What is 0.135 as a repeating decimal to a fraction?
1 answer:
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Short Answer
There are two numbers
x1 = -0.25 + 0.9682i <<<< answer 1
x2 = - 0.25 - 0.9582i <<<< answer 2
I take it there are two such numbers.
Let one number = x
Let one number = y
x + y = -0.5
y = - 0.5 - x (1)
xy = 1 (2)
Put equation 1 into equation 2
xy = 1
x(-0.5 - x) = 1
-0.5x - x^2 = 1 Subtract 1 from both sides.
-0.5x - x^2 - 1 = 0 Order these by powers
-x^2 - 0.5x -1 = 0 Multiply though by - 1
x^2 + 0.5x + 1 = 0 Use the quadratic formula to solve this.
a = 1
b = 0.5
c = 1
x = [-0.5 +/- sqrt(0.25 - 4)] / 2
x = [-0.5 +/- sqrt(-3.75)] / 2
x = [-0.25 +/- 0.9682i
x1 = -0.25 + 0.9682 i
x2 = -0.25 - 0.9682 i
These two are conjugates. They will add as x1 + x2 = -0.25 - 0.25 = - 0.50.
The complex parts cancel out. Getting them to multiply to 1 will be a little more difficult. I'll do that under the check.
Check
(-0.25 - 0.9682i)(-0.25 + 0.9682i)
Use FOIL
F:-0.25 * -0.25 = 0.0625
O: -0.25*0.9682i
I: +0.25*0.9682i
L: -0.9682i*0.9682i = - 0.9375 i^2 = 0.9375
Notice
The two middle terms (labled "O" and "I" ) cancel out. They are of opposite signs.
The final result is 0.9375 and 0.0625 add up to 1
There are two numbers
x1 = -0.25 + 0.9682i <<<< answer 1
x2 = - 0.25 - 0.9582i <<<< answer 2
I take it there are two such numbers.
Let one number = x
Let one number = y
x + y = -0.5
y = - 0.5 - x (1)
xy = 1 (2)
Put equation 1 into equation 2
xy = 1
x(-0.5 - x) = 1
-0.5x - x^2 = 1 Subtract 1 from both sides.
-0.5x - x^2 - 1 = 0 Order these by powers
-x^2 - 0.5x -1 = 0 Multiply though by - 1
x^2 + 0.5x + 1 = 0 Use the quadratic formula to solve this.
a = 1
b = 0.5
c = 1
x = [-0.5 +/- sqrt(0.25 - 4)] / 2
x = [-0.5 +/- sqrt(-3.75)] / 2
x = [-0.25 +/- 0.9682i
x1 = -0.25 + 0.9682 i
x2 = -0.25 - 0.9682 i
These two are conjugates. They will add as x1 + x2 = -0.25 - 0.25 = - 0.50.
The complex parts cancel out. Getting them to multiply to 1 will be a little more difficult. I'll do that under the check.
Check
(-0.25 - 0.9682i)(-0.25 + 0.9682i)
Use FOIL
F:-0.25 * -0.25 = 0.0625
O: -0.25*0.9682i
I: +0.25*0.9682i
L: -0.9682i*0.9682i = - 0.9375 i^2 = 0.9375
Notice
The two middle terms (labled "O" and "I" ) cancel out. They are of opposite signs.
The final result is 0.9375 and 0.0625 add up to 1
Answer:
3,750 cars.
Step-by-step explanation:
We are given that the equation:
Models the relationsip between <em>y</em>, the number of unfilled seats in the stadium, and <em>x</em>, the number of cars in the parking lot.
We want to determine the number of cars in the parking lot when there are no unfilled seats in the stadium.
When there are no unfilled seats in the stadium, <em>y</em> = 0. Thus:
Solve for <em>x</em>. Subtract 9000 from both sides:
Divide both sides by -2.4:
So, there will be 3,750 cars in the parking lot when there are no unfilled seats in the stadium.