CalculatorUser

There should be a filter for 'troll' and all the posts deemed as troll posts will pop up lol

\(\frac{n!}{r!(n-r)!}\) is the formula you always use in these types.

N represents number of items you are choosing out of, and r represents the number of items you are choosing.

Sorry! Here I will solve it myself.

2 cases here

case 1 : 1 topping

case 2 : 2 toppings

In all there should be 106 - 1 (no toppings case) = 105 ways with 2 cases added together.

OK....

Notice how if you have x toppings with y ways there are y/x different toppings.

So we have 1 topping and 2 toppings with 105 ways. That means there are 35 different toppings

This makes sense. Because :

CASE 1 : 1 * 35 = 35 ways

CASE 2 : 2* 35 = 70 ways

CASE 3 : Not toppings (1)

So add three cases together 35 + 70 + 1 = 106

So there you have it! :D :D :D

When looking at systems of linear equations. You can either use substitution or elimination.

The question is WHICH one to use? 

Don't skim my answer for the answer

(ax with a being the constant)

-----------

Substitution. You should use this whenever you see a variable with a constant of 1. For example the second equation you posted.

3x - y = 4

y = 2x - 2 

Notice how this can EASILY be substituted

Solving:

3x - (2x-2) = 4  (remember there is a -1 multiplying at the parentheses).

(remove parentheses and combine like terms.)

x + 2 = 4

x = 2

Now substitute back into one of your equations and find y.

Elimination. You should use this whenever you can't divide both sides of one of the equations and get integer constants.

4x + 7y = 6

9x - 2y = 49

Notice how the equation will be difficult substituting. If you try to isolate a variable, you will get fraction constants.

So a good strategy is to find the greatest common multiple.

Solving:

4x + 7y = 6            (Mulitply 7y by 2)

9x - 2y = 49           (Multiply 2y by 7)

8x + 14y = 12

63x - 14y = 343

Now you can just add both equations: 71x = 355

Then divide - x=5

Now substitute back into one of the equations and find y

There are 2^10 different subsets in the original set.

Now let us just subtract subsets without a 5

Let us just pretend you have set without a 5

So {1 2 3 4 6 7 8 9 10} Is your new set

So how many subsets can you make from that?

2^9 subsets

2^10 - 2^9 = 512 different subsets

NOTE: 2^10 - 2^9 is NOT 2.

So there you have it! :D:D:D

I am a middle schooler who likes math a lot, so I just learn as much as I can.

On an online calculator, that would work.

I don't have time so I'm going to answer the last one

If you are to choose 4 out of x students it will be \(\frac{x!}{4!(x-4)!}=y\)

Then if the number of students in the group increases by 1, the committees double.

So \(\frac{(x+1)!}{4!((x+1)-4)!}=2y\)

So we have two equations with 2 variables.

You can try and solve now, (try substituting)

Ok, when there is a geometric sequence, each TERM is multiplied by the common ratio to get the NEXT TERM.

So you can make the common ratio 5, for example.

Geometric sequence: She starts with 1 penny, and adds five times her current amount to her penny collection every day.

How many pennies she would have after 10 days?

Formula for sum of finite geometric sequence \(S_n=\frac{a(r^n-1)}{r-1}\)

Plugging in what we know - 

n = 10 (term)

r = 5 (common ratio)

a = 1 (first term)

9765624/4 = 2441406 pennies.