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3 years ago

How can you determine whether a counting number is a composite

1 answer:

4 0

Answer:A composite number is divisible by at least three numbers. Every counting number except 1 is either prime or composite. The reason 1 is neither is that it's divisible by only one number, which is 1. Remember the first four prime numbers: 2, 3, 5, and 7.

Step-by-step explanation:cant right no cause im in school but like and rate please.

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I need solving 4.8 (x+4)=2.6

victus00 [196]

<span>4.8 (x+4)=2.6
Use distributive property
4.8x + 19.2= 2.6
Subtract 19.2 from both sides
4.8x = -16.6
Divide 4.8 on both sides so that the only thing remaining on the left side is the variable x.
Final Answer: x = -3.458</span>

7 0

3 years ago

Read 2 more answers

How do you write 3/4 as percentage?

german

3/4 is 75%

for this think of them as quarters 4 quarters is 100 %, 2 quarters is 50% so 3 quarters would be 75%

or you could just do 3 divided by 4 then times that by 100

5 0

3 years ago

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Please help for brainiest; find values

love history [14]

Answer:

arc AB = 49°

arc ABC = 253°

arc BAC = 156°

arc ACB = 311°

Step-by-step explanation:

arc AB sees the center angle and it is equal to the center angle it sees

arc ABC is equal to 360 - 107 = 253° because it is also sees center angle

arc BAC is equal to 107 + 49 = 156°

arc ACB is equal to 360 - 49 = 311°

4 0

3 years ago

Good points... just answer the question correctly

NeTakaya

Answer:

R'(-9,4) --------------- R'(9,-4)

Step-by-step explanation:

5 0

3 years ago

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Solve the equation x^6+3x^5+x^4-5x^3-6x^2-2x=0 in the real number system

koban [17]

First, pull out a factor of $x$ .

$x^6+3x^5+x^4-5x^3-6x^2-2x=x(x^5+3x^4+x^3-5x^2-6x-2)$

Notice that when $x=-1$ (which you can arrive at via the rational root theorem), you have

$(-1)^5+3(-1)^4+(-1)^3-5(-1)^2-6(-1)-2=-1+3-1-5+6-2=0$

which means you can pull out a factor of $x+1$ . Upon dividing you get

$\dfrac{x^5+3x^4+x^3-5x^2-6x-2}{x+1}=x^4+2x^3-x^2-4x-2$

The rational root theorem will come in handy again, suggesting that $x=-1$ appears a second time as a root, which means

$\dfrac{x^4+2x^3-x^2-4x-2}{x+1}=x^3+x^2-2x-2$

Now this is more readily factored without having to resort to the rational root theorem. You have

$x^3+x^2-2x-2=x^2(x+1)-2(x+1)=(x^2-2)(x+1)$

so in fact, $x=-1$ shows up as a root for a third time.

So, you have

$x^6+3x^5+x^4-5x^3-6x^2-2x=x(x+1)^3(x^2-2)=0$

Two roots are obvious, $x=0$ and $x=-1$ (with multiplicity 3). The remaining two are given by

$x^2-2=0\implies x^2=2\implies x=\pm\sqrt2$

3 0

4 years ago