Answer:
Number line drawing attached.
Step-by-step explanation:
Simplify each fraction:
1/6 is simply 1/6
2/6 is the same as 1/3
3/6 is the same as 1/2
4/6 is the same as 2/3
5/6 is simply 5/6
6/6 is the same as simply 1
20% of 100 is 20it's not that hard just don't be lazy
Answer:
x = (i sqrt(7))/4 - 3/4 or x = -(i sqrt(7))/4 - 3/4
Step-by-step explanation:
Solve for x:
x - 1 - 2/x = 3 x + 2
Bring x - 1 - 2/x together using the common denominator x:
(x^2 - x - 2)/x = 3 x + 2
Multiply both sides by x:
x^2 - x - 2 = x (3 x + 2)
Expand out terms of the right hand side:
x^2 - x - 2 = 3 x^2 + 2 x
Subtract 3 x^2 + 2 x from both sides:
-2 x^2 - 3 x - 2 = 0
Divide both sides by -2:
x^2 + (3 x)/2 + 1 = 0
Subtract 1 from both sides:
x^2 + (3 x)/2 = -1
Add 9/16 to both sides:
x^2 + (3 x)/2 + 9/16 = -7/16
Write the left hand side as a square:
(x + 3/4)^2 = -7/16
Take the square root of both sides:
x + 3/4 = (i sqrt(7))/4 or x + 3/4 = -(i sqrt(7))/4
Subtract 3/4 from both sides:
x = (i sqrt(7))/4 - 3/4 or x + 3/4 = -(i sqrt(7))/4
Subtract 3/4 from both sides:
Answer: x = (i sqrt(7))/4 - 3/4 or x = -(i sqrt(7))/4 - 3/4
These are the answer that I got:
A. <span>That means: "The more it costs, the fewer people you can invite."
B. <span>Amount you can spend = M; You will have to come up with this number - whatever you think is reasonable.
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C. (C = Cost per person, P = number of people you can invite )
D. You will have two equations. In the first equation, C will be the per-person cost at the first bowling alley. In the second equation, C will be the per-person cost at the second bowling alley.
E. <span>Be sure to think this through. Yes, you will be able to host more guests at the bowling alley with the lowest per-person cost, but is that the only factor you need to consider? What about location? Quality of lanes and bowling equipment? Music? Food? Customer Service? Consider ALL aspects.</span> You might pick the more expensive bowling alley because you feel would have a better experience, or maybe you don't want to invite lots of people anyway.<span>
Hope this helps!</span>
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