Welcome back EP !!.
See diagram. x = side length
0 .000 000 000 000 126 139 935
Ok it was just 6.2 nevermind!
same
Mondays...
sorry to i like donuts he got hacked and the hacker said some mean stuff but its good
1. B
2. B
8)
The one selected in yellow
the height of an equilateral triangle is
\(x*\sqrt3\)
and the length of a side is
\(2x\)
seeing as they both have x, the side length is
\(2*127\)
or 254
aha thanks
Come on Ginger, there is no need to be mean
I can see your answer is correct but our guest's error is easily made.
Brad can type 1260 words in an hour. What is Brad’s average typing rate in words per minute?
Season’s greets, Mr. BB:
‘Tis the season to . . .
Deck the halls with boughs of BLARNEY Troll-la-Troll-la-Troll, la-Troll-Troll, Troll.
There is nothing wrong with this Yule time question. It does require advanced, grade-school mathematics skills to answer it, though.
\(500000 + \underbrace{ (200*300)}_{\scriptsize \text{Portion tied to profits}} + (17*500) = 568500 \\ \dfrac{(200*300)}{568500} = 0.1055 =10.6 \mathbf{\%} \)
GA
Thanks Tiggsy
What is the second value of b? Is it 2 or 3 ?? Look at the title and the body of the question !
At any rate, I believe it goes like this:
a = k x 1/b^2, where k = constant
9 = k x 1/2^2
k= 36
a = 36 x 1/4 =9 when b=2, or
a = 36 x 1/9 =4 when b=3
I think questions lock after about a week.
If you want to add to them just send a moderator the address by private message and we can unlock the question for you.
How could the CEO's salary be: W = 500,000 + 200 π + 17S
=500,000 + [200*$300,000,000] + [17*$500,000,000]
=$68,500,500,000 !!!!!! Do you see the mistake????
Thanks m8
See the answer here:
http://web2.0calc.com/questions/how-many-ways-are-there-to-put-4-b-s-in-3-boxes-if-the-b-s-are-distinguishable-but-the-boxes-are-not
I am sorry, I misread the question. I thought there were 4 boxes.
Thank you, but that is incorrect.
Solution I found:
We will consider this as a composite of two problems with two indistinguishable balls and 3 distinguishable boxes. For two indistinguishable green balls, we can place the balls in a box together or in separate boxes. There are 3 options to arrange them together (in box 1, 2, or 3) and 3 options for placing them separately (nothing in box 1, 2, or 3). So there are 6 ways to arrange the indistinguishable green balls. By the same reasoning, there are 6 ways to arrange the indistinguishable red balls, for a total of 6*6=36 arrangements of the 4 balls.
thank you guys!
The Argand diagram is the complex plane.
Thank you so much!
