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(10/3 )^2 = 11.11     and    (42/5)^2 = 70.56

   so all integers from 12 to 70 are included....    You can finish

corrections:

A = (3√3) / 2 * a²  

Since any three points determine a plane, the answer is $\binom{8}{3} = \boxed{56}$.

Here's your answer:

\(\frac{y}{3}+1=\frac{y+3}{y}\)

\(y^2+3y=3\left(y+3\right)\)                          (multiply both sides by 3y)

\(y^2+3y=3y+9\)                              (distribute right side)

\(y^2=9\)                                                 (subtract both sides by 3y)

\(y=3,\:y=-3\)                                  (final answer)

#!: Split the hexagon into 16 parts. 10 are shaded. 10/16 = 5/8. 5 + 8 = 13

#2: Split the hexagon like this: 

So each of the triangles is 60/8 = 7.5,  and 4 * 7.5 = 30. 60 + 30 = 90

Hexagon area:

Hexagon side => a

a(a√3) = 60          a = sqrt(60 / √3)

A = (3√3)/2 * a

The problem is fairly straight-forward if you use inversion.

By the quadratic formula, t = 2.1 seconds.

ita not 3. it says You're right that 3 satisfies the equation. But are there any other values of y

srry it does not show up

y/3 + 1 = (y + 3)/y

==> (y + 3)/3 = (y + 3)/y

==> y = 3

The only solution is y = 3.

64^2 = 4096

45*2 = 90

New simplified equation is, 4096 + 145 - 90 - 14 = 4137

This equals (2^3)^2 - (3^2)^3 = 2^6 - 3^6 = 64 - 729 = -665. 

Hey there, ICL!

So, sometimes when you can't solve an equation, you can plug it into a Desmos Graphing Calc and then you'll be able to see the vertex.

EP is right about this one, x=2 and y= -7; here's the link to the graph: https://www.desmos.com/calculator/9nnq18slcn.

Hope this helped! :)

( ﾟдﾟ)つ Bye

First lets choose the stars on the court. 

3 choose 2 = 3 * 2/2 * 1 = 3

Now, out of the 8 players left, 3 of them need to be on the court. 

8 choose 3 = 8 * 7 * 6/6 = 56

56 * 3 = 168

There are 168 possibilities. 

Hey there, Guest!

So, the equation is \(\left(a^b\right)^a-\left(b^a\right)^b\ \) and we'll substitute properly.

\(\left(2^3\right)^2-\left(3^2\right)^3\)

\(8^2−(3^2)^3\)

\(64−(3^2)^3\)

\(64−9^3\)

\(64−729\)

= −665

Hope this helped! :)

( ﾟдﾟ)つ Bye

The number of total arrangements is 8!/4 * 2! = 8! / 8 = 7! = 5040. 

There are 5040 total arrangements.

The number of arrangements in which the first four balloons have different colors is 4! * 4! = 24 * 24 = 576. 

There are 576 total arrangements.

Hope this helped!!

There are two cases,

1. Two all stars players and picked and there are 3 regular players

2. Three all star players are picked and there are 2 regular players

Let's start with case 1. There are 3c2 ways to pick the 2 all star players that will play, 3 ways. And there are 7c3 ways to pick the 3 regular players that play, 35 ways. 3*35 = 105, so 105 possible ways for case one.

For case 2, it is exactly the same logic, just different values. 3c3 = 1 way, and 7c2 = 21, 21*1 = 21.

21+105 = 126 ways i think

We can use PIE, (Principle of Inclusion - Exclusion)

We can add 48 + 39 + 35 = 122, but this is over counting since some people took 2 or even 3 classes. 

From 122, we need to subtract 20 + 19 + 22 = 61.

But we need to add 10 back to count the ones who took all the classes. 

61 + 10 = 71. 

100 - 71 = 29

29 students did not enroll in any of the subjects. 

Substitute both numbers in ===> (2^3)^2 - (3^2)^3 = 2^6 - 3^6 = -665

i figured out myself...answer is 11C3...but thanks for trying to help! 

Out of the 18 people, 2 of them will be on the decoration committee. 

18 choose 2 = 18 * 17 / 2 = 9 * 17 = 153

Then out of the 2 people chosen, on will be the leader. 

2 choose 1 = 2/1 = 2

153 * 2 = 306

There are 306 ways the committee can be choosen.