Answer:
There is no solution to the systems of equation.
Step-by-step explanation:
Graph the system by using y=mx+b
Both systems are y=2/5x+5/2.
10000 digits can be used for 4 digit A.T.M code.
<u>Solution:</u>
Given that A.T.M required 4 digit codes using the digits 0 to 9.
Need to determine how many four digit code can be used.
We are assuming that number starting with 0 are also valid ATM codes that means 0789 , 0089 , 0006 and 0000 are also valid A.T.M codes.
Now we have four places to be filled by 0 to 9 that is 10 numbers
Also need to keep in mind that repetition is allowed in this case means if 9 is selected at thousands place than also it is available for hundreds, ones or tens place .
First digit can be selected in 10 ways that is from 0 to 9.
After selecting first digit, second digit can be selected in 10 ways that is 0 to 9 and same holds true for third and fourth digit.
So number of ways in which four digit number is created = 10 x 10 x 10 x 10 = 10000 ways
Hence 10000 digits can be used for 4 digit A.T.M code.
The square root of -5
2.23606798
or
-2.23606798
Answer:
6^ -24
Step-by-step explanation:
We know that a^b^c = a^ (b*c)
(6^-4)^6 = 6^ (-4*6) = 6^ -24
Answer:
8 intersections
Step-by-step explanation:
Miss Stevens drove through a total of 36 intersections on her way home from work last week.
At four of every 16 intersections Miss Stevens had to stop for a red light before she could drive through .
This is calculated as:
36 intersections ÷ 16 intersections
= 2 4/16
= 2 1/4
At four of every 16 intersections Miss Stevens had to stop for a red light before she could drive through .
4 of every 16 intersections = stop for a red light
Hence:
16/4 = 4 intersections
Hence: 4 intersections = 1 stop for a read
The number of intersections that Miss Stevens had to stop for a red light is:
2 × 4 intersections = 8 intersections
