Answer:
Good luck honey.
Step-by-step explanation:
Everyone on this website don't like to help other people who need answers quick. I would help you but the pictures are blocked because i have a shcool computer. type the question out and will help. maybe someone will eventually help me.
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
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Answer:
i cant see it...........
Step-by-step explanation:
Answer:
The other diagonal measures 21m
Step-by-step explanation:
In this question, we are tasked with calculating the length of the second diagonal of as Rhombus given the measure of the surface area of the rhombus and the length of the other diagonal
Mathematically, for a rhombus having two diagonals 
and 
, the area of the rhombus can be calculated mathematically using the formula below;
A = 1/2 × ×
From the question, we can identify that A = 157.5
and = 15m
we input these in the formula;
157.5 = 1/2 × 15 ×
315 = 15
= 315/15
= 21m
Alex did the most amount of the homework assignment because if you turn them all into a fraction with a common denominator, Alex would have the most.
The fractions will be 80/120, 60/120, 60/120, and 105/120 which is Alex.
