Q1. The answer is 4(2x - 3)(2x + 1)
16x² – 16x – 12 = 4 * 4x² - 4 * 4x - 4 * 3 =
= 4(4x² - 4x - 3) =
= 4(4x² + 2x - 6x - 3) =
= 4(2x * 2x + 2x - (2x * 3 + 3)) =
= 4(2x(2x + 1) - (3(2x + 1))) =
= 4((2x + 1)(2x - 3)) =
= 4(2x - 3)(2x + 1)
Q2. The answer is 3(x + 8)(x - 1)
3x² + 21x – 24 = 3 * x² + 3 * 7x - 3 * 8 =
= 3(x² + 7x - 8) =
= 3(x *x - x + 8x - 8) =
= 3((x(x - 1) + 8(x - 1)) =
= 3(x + 8)(x - 1)
The answer would be f(5)=5x+15 I think.
0 and 1 are neither prime nor composite. A prime is any number greater than 1 that has just 1 and itself as factors. Primes can only start at x > 1
When that happens (when you start with numbers greater than one) p^2 is a composite consisting of 2 primes, so any composite will obey the law that he number will have at least 3 factors making it up -- in this case p p^2 and 1.
So the answer to the question by definition is that 0 numbers can have the property of both p and p^2 to be prime.
Answer:
a) Y = -3x/5 + 15
b) Y = 3 - 0.6x
d) Volume = 15Pi
e) Volume of cone = 15pi
Step-by-step explanation:
The detailed steps and appropriate integration is as shown in the attached file.
