Cole said All prime numbers are odd and all composite numbers are even do you agree with cole?

2 answers:

4 0

All prime numbers are odd except for the number two. A prime number is defined as any whole number greater than one that has no positive divisors except for one and itself. Since two can only be divided by the numbers one and two, it is prime. The composite numbers are integers having at least two prime factors, whereas a prime number has a single prime factor (itself), and the unit 1 has no prime factors at all. For example, 10 is composite, as it has two prime factors (2 and 5). 11 is notcomposite, since its only prime factor is itself.

Anit [1.1K]3 years ago

3 0

A prime number is such that it is divisible by only itself and one. ... Two is a prime because it is divisible by only two and one. All the other even numbers are not prime because they are all divisible by two. That leaves only the odd numbers.

You might be interested in

$10.94 5% sales tax how much is the total

Dmitrij [34]

Find 5 percent of 10.94, which is .55. Since it is sales tax add .55 to 10.94 You get 11.49! Hope this helps

3 0

3 years ago

Read 2 more answers

Forth grade has collected 2000 boxtops. Fifth grade has fifty more how many more does fifth grade have

Law Incorporation [45]

Answer:

2050

Step-by-step explanation:

2000+50=2050

5 0

3 years ago

Read 2 more answers

Write the equation of the line that passes through the points (2,-6) and (6,-6).

NeTakaya

Answ

Step-by-step explanation:

Two point slope =-(-6+6)/(6-2) =0

Equation is y+6 = 0(x-2)

y+6=0

y=-6

The point is -6 at y axis which is a horizontal line.

5 0

2 years ago

Express the complex number in trigonometric form. -6 + 6 square root of 3 i ?

bagirrra123 [75]

$\bf \begin{array}{clclll}
-6&+&6\sqrt{3}\ i\\
\uparrow &&\uparrow \\
a&&b
\end{array}\qquad 
\begin{cases}
r=\sqrt{a^2+b^2}\\
\theta =tan^{-1}\left( \frac{b}{a} \right)
\end{cases}\qquad r[cos(\theta )+i\ sin(\theta )]\\\\
-------------------------------\\\\$

$\bf r=\sqrt{(-6)^2+(6\sqrt{3})^2}\implies r=\sqrt{36+(6^2\cdot 3)}\implies r=\sqrt{144}
\\\\\\
r=12\leftarrow 
\\\\\\
\theta =tan^{-1}\left( \frac{6\sqrt{3}}{-6} \right)\implies \theta =tan^{-1}\left( -\sqrt{3}\right)\implies \theta =
\begin{cases}
\frac{2\pi }{3}\leftarrow \\
\frac{5\pi }{3}
\end{cases}$

now, notice, there are two valid angles for such a tangent, however, if we look at the complex pair, the "a" is negative and the "b" is positive, that means, "x" is negative and "y" is positive, and that only occurs in the 2nd quadrant, so the angle is in the second quadrant, not on the fourth quadrant.

thus $\bf 12\left[cos\left( \frac{2\pi }{3} \right) +i\ sin\left( \frac{2\pi }{3} \right) \right]$

5 0

3 years ago

Only 8 9 and 10 please

sattari [20]

8) 3/16
9) 41/8
10) 218/5

3 0

3 years ago