To find this, first find the factor or rate of which the numbers are moving. To do so do as follows.
subtract 1 from 3
3-1=2
So each number is having 2 added to it.
Now add two to 7 and the numbers afterwards till you get the 12th term
7+2=9
1+3+5+7+9
9+2=11
1+3+5+7+9+11
11+2=13
1+3+5+7+9+11+13
13+2=15
1+3+5+7+9+11+13+15
15+2=17
1+3+5+7+9+11+13+15+17
17+2=19
1+3+5+7+9+11+13+15+17+19
19+2=21
1+3+5+7+9+11+13+15+17+19+21
21+2=23
1+3+5+7+9+11+13+15+17+19+21+23
So 23 is the 12th term
I'm not sure I'm understanding the wording of the question, but if it's this:
Juice boxes come in a package with multiple juice boxes in each package. Three people bought 18, 36, and 45 juice boxes. What is the largest possible number of juice boxes per package?
Then the problem is just an involved way of asking what the greatest common factor of 18, 36, and 45 is, and the answer is 9, the difference between 36 and 45, which are both multiples of 9. Note that 18 is also a multiple of 9. One way to find the greatest common factor of three numbers is to factor all of them and find which prime factors they have in common.
Answer:
The number x is -3
Therefore x=-3
Step-by-step explanation:
Given problem is "Three added to eight times a number is the same as three times the value of two times the number -1"
Let x be the number
It can be written as
( applying the distributive property )
subtracting (6x-3) on both sides we get
( ading the like terms )
Therefore x=-3
The number x is -3
f(h(x))= 2x -21
Step-by-step explanation:
f(x)= x^3 - 6
h(x)=\sqrt[3]{2x-15}
WE need to find f(h(x)), use composition of functions
Plug in h(x)
f(h(x))=f(\sqrt[3]{2x-15})
Now we plug in f(x) in f(x)
f(h(x))=f(\sqrt[3]{2x-15})=(\sqrt[3]{2x-15})^3 - 6
cube and cube root will get cancelled
f(h(x))= 2x-15 -6= 2 x-21
