there are 24 marbles in a bag: 2/16 of the marbles are red, 1/8 of the marbles are blue, 1/4 of the marbles are yellow, 1/3 of the marbles are black, 1/6 of the marbles are silver. how many of each type of marble is there?

3 red marbles, 3 blue marbles, 6 yellow marbles, 8 black marbles, 4 silver marbles

Step-by-step explanation:

I know these are the correct number of marbles because I set up these numbers as fractions and solved for x.

For examples, 2/16 of the marbles are red. 2/16=x/24

You would do this because we are trying to find the number of marbles that are red out of the 24 marbles.

Next, we would do 24 divided by 16= 1.5 Then 1.5 times 2=3

You would follow this rule for each fraction of marbles in order to find the true number of marbles out of 24.

8

step-by-step explanation:

to find the amount of gift bags to use all the supplies, find the gcf or greatest common factor. this is the largest number which multiplies evenly into each term.

48 has factors: 1, 2, 3, 4, 6, 8, 12. 16, 24, 48

80 has factors: 1, 2, 4, 5, 8, 10, 16, 20, 40, 80

64 has factors: 1, 2, 4, 8, 8, 16, 32, 64

the highest number factor they all share in common is 8. this means there will be 8 bags with 6 pencils, 10 markers, and 8 stickers in each bag.

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!