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.
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Answer:
Step-by-step explanation:
I am assuming that by "interior" angle you do not mean the central angle.
This is a 10-sided polygon, a decagon. That means that there are 10 triangles that can extend from the center, with their sides being equal to the radii of the decagon. If we extract one of these triangles we can find what the interior angle is. The vertex angle measures 360/10 which is 36.
Split this triangle in half from the vertex to the base, creating a right triangle. The vertex angle is also split in half, making this angle (the vertex angle is the one at the top of the triangle) 18 degrees. We already know that one angle inside this right triangle is 90 (definition of a right triangle) and to find the other one, we apply the Triangle Angle-Sum Theorem:
180 - 18 - 90 = 72 degrees. That is the measure of the base angle that is NOT the right angle, obviously.
It is 6 was not multiplied by 3
The answer should be 6(m+3)= 6m+ 18
First, let's fine the decimal of each fraction.
-5 Divided By 6 = -0.83333333333
17 Divided By 18 = 0.94444444444
-2 Divided By 9 = -0.22222222222
-5/6 - 17/18 - (-2/9) = -1.55555555556
