Answer:
Johnny is correct because irrational numbers never end
Well, you could assign a letter to each piece of luggage like so...
A, B, C, D, E, F, G
What you could then do is set it against a table (a configuration table to be precise) with the same letters, and repeat the process again. If the order of these pieces of luggage also has to be taken into account, you'll end up with more configurations.
My answer and workings are below...
35 arrangements without order taken into consideration, because there are 35 ways in which to select 3 objects from the 7 objects.
210 arrangements (35 x 6) when order is taken into consideration.
*There are 6 ways to configure 3 letters.
Alternative way to solve the problem...
Produce Pascal's triangle. If you want to know how many ways in which you can choose 3 objects from 7, select (7 3) in Pascal's triangle which is equal to 35. Now, there are 6 ways in which to configure 3 objects if you are concerned about order.
5/6 of the paper is devoted to news and pictures
first I made the fraction have equal denominators which was 6 by doing that I multiplyed the numerator and the denominator
1/2
1×3=3
2×3=6
therefore 1/2 being equivalent to 3/6
1/3
1×2=2
3×2=6
therefore 1/3 being equivalent to 2/6
now you would be able to add the numerators, so 2+3=5
therefore being 5/6 of the paper is devoted to news and pictures
Answer:
x = 7/2 or x = -8/3
Step-by-step explanation:
Solve for x over the real numbers:
6 x^2 - 5 x = 56
Divide both sides by 6:
x^2 - (5 x)/6 = 28/3
Add 25/144 to both sides:
x^2 - (5 x)/6 + 25/144 = 1369/144
Write the left hand side as a square:
(x - 5/12)^2 = 1369/144
Take the square root of both sides:
x - 5/12 = 37/12 or x - 5/12 = -37/12
Add 5/12 to both sides:
x = 7/2 or x - 5/12 = -37/12
Add 5/12 to both sides:
Answer: x = 7/2 or x = -8/3
F - g(2) = 3x^2 + 1 - ( 1 - x )
= 3x^2 + 1 - 1 + x
= 3x^2 + x
= 3(2)^2 + 2
= 14
the answer is B.14
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