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The spinner has 5 sides. Therefore (assuming the sides are all different, which the problem should've noted), there are 5 outcomes when you roll the spinner. When you flip a fair coin (notice the problem should've noted the coin was fair), there are always 2 outcomes (heads/tails). Therefore, by the Multiplication Principle, (look that up if you don't know what it is) the answer is 5 * 2  = 10

Hope this helped!

Those weren't right. Could you also explain a bit? Thanks

Hmmm..

Yay now I figured it out!!!!

We know that a/b = r

Then a = rb

so a + b = rb + b = b(r + 1)

a - b = rb - b = b(r - 1)

Therefore the ratio is (b(r+1))/(b(r-1))

This is just (r+1)/(r-1)

And that's the answer!!!!!

I tries this answer and it is incorrect. Thank you though.

You can take the vectors [3,4], [4,1], and [5,-2].

There is a circle being inscribed in an equilateral triangle, such that it touches all 3 sides of the triangle.  What is the ratio of the radius of the circle to the side length of the triangle?

thank you!

Using Quadratic Formula , you will find   roots  x =150/36   x=−192/36   Square then  add them for your answer....

ABCDE is a regular pentagon.  On the outside of AB, we construct a square ABFG.  What is the measure, in degrees, of angle EAG?

Thanks for the information. I have corrected my mistakes.
I'm sorry.

As an improper fraction, for how long is the cannonball above a height of 6 meters?

Find the equation whose graph is a parabola with vertex (2,4), vertical axis of symmetry, and contains the point (1,1). Express your answer in the form "ax^2+bx+c". Thanks!

I actually figured it out a while after I posted this question haha. My process was the exact same as yours, though

Wow thanks a lot Eletric Pavlov you are very smart

Here is a way:

Vertex value of 'x' is at   -b/2a     soooo     -b/2a = 2  (1) 

                                                  and when x = 2        f(x)=4

                                                            4 = a(2^2)+b(2) + c   or  4 = 4a+2b+c         (2)

    and you are given point 1,1            1 = a(1^2)+b(1)+c     or   1 = a+b+c            (3)       3 equations , 3 unknowns

re-arrange (1) :   -b/2 = 2a      multiply by 2

                               -b =4a          sub into (2)

 4 = -b+2b+c    4 = b+c     c = 4-b

Take (3) and multiply by 4     4 = 4a + 4b + 4c  

                                             4 = -b+4b + 4(4-b)    so b = 12    then     -b/2a=2 means   a = -3      then a + b + c =1 means c = -8

The value of k is 5.

The smallest number I could find is:

123,475,869 mod 99 = 0

Using casework, there are C(7,3) + C(5,3) + C(3,3) = 35 + 10 + 1 = 46 ways to distribute the candies.

 n=32; k=6; p=0;a= n*k nCr 6;p=p+a;n--;k++;if(k<=37, goto3,0);printp;a=10; b=1;c= a nCr b; if(c==61523748, goto4, goto5);printc, a, b; a++;if(a<100, goto2, 0);a=10;b++;if(b<100, goto2, discard=0;

Sum =61,523, 748 

n = 39   and   k = 8, and 

n = 39   and   k =31

First, notice that if all the candies were different, it would simply be 4^15 (4 possible kids for each candy). If all the candies were the same, it would be (4 multichoose 15). This problem is something in between.

The answer is (4 multichoose 8) * (4 multichoose 7) = C(11,8)*C(10,7) = C(11,3)*C(10,3) = 19800

Here is the last time this was asked.  (and answered)

Using the Law of Cosines, we can find the side of the octagon, S, as

S^2  =  1^2 + 1^2   - 2 ( 1 * 1) cos (45°)

S^2   = 2  - [ 2 *  √2 / 2 ] 

S^2  = 2 - √2

S = √ [ 2 - √2 ]    units

So....the perimeter is

8 √ [ 2  - √2 ] units    ≈  6.12 units

(a + bi)^3  =

a^3  + 3a^2 (bi)  +  3a (bi)^2 + (bi)^3

a^3  + 3a^2bi  - 3ab^2 - bi

We need to  have

3a^2bi  - bi  =   107i

bi ( 3a^2 - 1)  =  107i                divide both sides by i

b ( 3a^2 - 1)  = 107

Since 107 is prime.....then....

Let  b   = 1

So

3a^2 - 1 = 107

3a^2  = 108

a^2 = 36

a = 6

So

(6 + i)^3    - 107i  =

6^3  + 3(6)^2 i   + 3(6)(i)^2  - i   - 107i   =

216 + 108i  - 18 - i  - 107i   =

(216 - 18) +  (108i - 108i)  =

198  =  the real number  

Nice, CU   !!!!!

A nunber is divisible by 99 if it is divisible by 9 and 11.

Divisibility rule for 9 is if all the numbers sum up to a multiple of 9, then the number is divisible by 9.

Any number made of digits 1 through 9 is divisilble by 9.

so basically the question is now:

Find the smallest number using digits 1-9 that is divisible by 11.

We start with 123456789. Because its the smallest.

Using the picture above, we get

1 + 3 + 5 + 7 + 9 = 25

2 + 4 + 6 + 8 = 20

25 - 20 = 5. Which is not divisible by 11.

So how can we manipulate the red and blue numbers to create a result that is divisble by 11?

Welp, guess and check...

Sorry, that is the best I can do. I was able to at least limit the number of guesses  at least....