6 phone numbers are possible for one area code if the first four numbers are 202-1
<u>Solution:</u>
Given that, the first four numbers are 202-1, in that order, and the last three numbers are 1-7-8 in any order
We have to find how many phone numbers are possible for one area code.
The number of way “n” objects can be arranged is given as n!
Then, we have three places which changes, so we can change these 3 places in 3! ways
Hence 3! is found as follows:
So, we have 6 phone numbers possible for one area code.
Hello!
For this problem we are given that quadrilateral ABCD is congruent to quadrilateral GJIH, meaning that all sides and angle measures will be equivalent to its corresponding side.
This means that to find 
, we can look at quadrilateral GJIH's corresponding side to quadrilateral ABCD's side AD, which is side GH, which has a value of 9.
This means that 9 should also be the side length of side AD, which we're given a value of 
.
Solve.
Hope this helps!
Answer: x=3-2y/3
Step-by-step explanation:
You just have to try to isolate one variable
Answer:
(3, 9)
Step-by-step explanation:
Set 3x and 2x+3 equal to each other forming an equation in x.
3x=2x+3
Solve for x.
x=3
Substitute the given value of x into the equation y=3x.
y=3x3
Solve for y.
y=9
(x, y) = (3, 9)
Hope this helps :)
The first thing you do is go two space forward like start at the o and move two spaces forward you put a decimal at 9 like this 0.90 and that's how you do it
