2 years ago

I am in a subset of rational numbers, I am not negative, I am a whole number, I am not a natural number, what number am I?

Mathematics

1 answer:

MakcuM [25]2 years ago

3 0

Answer:

Step-by-step explanation:

0 is considered a whole number, but it is not a natural number. The natural numbers start at +1 and go forward. ) is not positive or negative, but being that it is a whole number, it is also a rational number.

You might be interested in

What is the vertex of the parabola shown?

Ivenika [448]

10,800 would be the best awnser for you

3 0

2 years ago

Read 2 more answers

Help me pls, math is my weak point TwT

Alja [10]

Answer:

I hope this helps

8 0

2 years ago

(Q2) Determine the eccentricity, the type of conic, and the directrix for r = 9/1+cos theta

g100num [7]

Answer:

The answer is (d) ⇒ eccentricity = 6 , hyperbola , directrix x = 3/2

Step-by-step explanation:

∵ r = ek/(1 ± ecosФ)

∵ r = 9/(1 + 6cosФ)

∴ e = 6 ⇒ eccentricity = 6

∵ e > 1

∴ The conic is hyperbola

∵ ek = 9

∴ k = 9/6 = 3/2

∴ The directrix x = 3/2

3 0

2 years ago

Item 9<br> Use the table to write a proportion.

Fed [463]

Answer:

We need an attachment or more details to help you.

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Please help me ASAP I WILL GIVE U BRAINLIEST!

cestrela7 [59]

Answer:

$\sqrt{13}$

Step-by-step explanation:

Given a vector < x, y > then its magnitude is

$\sqrt{x^2+y^2}$

Given the directed line segment

P (0, 0 ) → Q (3, 2 ) then

PQ = q - p = (3, 2 ) - (0, 0 ) = < 3 - 0, 2 - 0 > = < 3 , 2 >

| PQ | = $\sqrt{3^2+2^2}$ = $\sqrt{9+4}$ = $\sqrt{13}$

8 0

2 years ago