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3 years ago

13

Which rational number also belongs to the set of natural numbers? Select one: a. 15 b. -12 c. 38 d. 0.1666...

1 answer:

erma4kov [3.2K]3 years ago

7 0

Answer:

Any number that belongs to either the rational numbers or irrational numbers would be considered a real number. That would include natural numbers, whole numbers and integers. There are two parts to this: the number has to belong to the set of whole numbers {0, 1, 2, 3, ...} and.

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Trey is solving his equation, and the last line of his work is "-2 = 2". What does this mean?

oksian1 [2.3K]

Answer:

it means that his equation is not true.

Step-by-step explanation:

-2=2 is obviously incorrect, so we must assume that the equation isn't true.

3 0

3 years ago

What is the word for a rational number whose cube root is a whole number

xxTIMURxx [149]

It's called perfect cubes

8 0

3 years ago

gladu [14]

Answer:

fyi b's answer has imaginary numbers in it...

Imaginary: 1 + $\frac{\sqrt{2i} }{2 }$

Imaginary: 1 - $\frac{\sqrt{2i} }{2 }$

Step-by-step explanation:

$2x^{2} - 4x -3 = 0$

$\sqrt{-4^{2} -4(2)(3)}$ = $\sqrt{-8}$ ... the negative root will produce imaginary solutions

8 0

2 years ago

Read 2 more answers

Simplify 7 - (-4) + 3(-6)

jeka94

Answer:

-7

Step-by-step explanation:

7 - (-4)= 11

+

3(-6)= -18

=-7

8 0

3 years ago

Read 2 more answers

Given the arithmetic sequence: u(1)= 124, u(2)= 117, u(3)= 110, u(4)=103,...

omeli [17]

We are given

arithmetic sequence: u(1)= 124, u(2)= 117, u(3)= 110, u(4)=103

so, first term is 124

u(1)= 124

now, we can find common difference

$d=u(2)-u(1)$

$d=117-124$

$d=-7$

now, we can find kth term

$u(k)=u(1)+(k-1)d$

now, we can plug values

and we get

$u(k)=124+(k-1)*-7$

$u(k)=124-7k+7$

$u(k)=131-7k$

u(k) must be negative

so,

$u(k)=131-7k$

$131-7k$

now, we can solve for k

$7k>131$

$k>18.714$

so, it's closest integer value is

$k=19$ ..............Answer

3 0

3 years ago