<h3>
Answer: B. All real numbers</h3>
Work Shown:
2(3c+3) < 6c+9
6c+6 < 6c+9 .... distribute
6c+6-6c <6c+9-6c .... subtract 6c from both sides
6 < 9 ... note the c terms go away entirely
We're left with a statement that is always true since 6 is always less than 9 regardless of what value of c we pick. So there are infinitely many solutions. Any real number can replace c to get a true statement.
Answer:
there are a total of 120 + 96 = 216, 5 digit numbers divisible by '3' that can be formed using the digits 0 to 5.
Find the third iterate, x3, of the function f(x) = 3x + 5 for an initial value of x0 = 1. <span>x1 = 3(1) + 5 = 8 ... x2 = 3(8) + 5 = 29 ... x3 = 3(29) + 5 = 92 </span>
The picture is difficult to read, but it looks like
∠ADB = 2x²
∠ACB = 10x
These two angles have equal measure, so you have
2x² = 10x
2x(x -5) = 0 . . . . . subtract 10x and factor
x = 0, or x = 5
The measure of arc AB is twice the measure of either inscribed angle, so would be
arc AB = 2×10x
= 2×10×5
= 100 . . . . . degrees
Arc AB is 100°.
