Answer:
7
Step-by-step explanation:
Answer:
B. Congruent - SSS
Step-by-step explanation:
Since, corresponding sides of both the triangles are congruent. Hence, both the triangles are congruent by SSS Postulate.
The largest possible last digit in the string of 2002 digits and number divisible by 19 or 31 is 9.
Given the first digit of a string of 2002 digits is 1 and the two digit number formed by consecutive digits within the string is divisible by 19 or 31.
We have to tell the last largest digit of such number.
Two digit numbers divisible by 19=19,38,57,76,95.
Two digit numbers divisible by 31=31,62,93,124
Number started with 1 =19
Last digit is 9
We have said that the number should be divisible by 19 or 31 not from both and started with 1.
Hence the largest possible last digit and number divisible by 19 or 31 in this string is 9.
Learn more about digits at brainly.com/question/26856218
#SPJ4
1) You must do division first. So divide 5/12 and get .41666666...
2) Then you add 15+5 and get 20.
3) Finally you subtract 20 and .41666666... and you get 19.5833333...
4) you can simplify 19.583333... to 19.584 if you need to.
X = 3/2 or 1 1/2
Both of the equations equal -10, therefor the equations are equal.