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3 years ago

Tell whether the statement "All integers are rational numbers" is true or false. Explain your choice.

1 answer:

3 0

True!!
Bcoz integer numbers are positive and negative whole numbers. Rational numbers ALSO include fractions and decimals...... eg: 10 is an integer which is also a rational number. but keep in mind all rational numbers are not integers. eg 10.1 is a rational number but since integers are only whole positive or negative numbers, it is not an integer.

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The planet mercury takes 56.6 earth days to rotate once around its axis. About how many times does it rotate around its axis dur

Kruka [31]

In 56.6 earth days , it rotates 1 time
so, in 1 earth day , it will rotate $\frac{1}{56.6}$ times

now, In a earth year, there are 365 days.

So, in 365 earth days, it will rotate = $\frac{1}{56.6} \times 365 = 6.44$

6 0

4 years ago

Isaiah is a college student who rents an apartment that costs $755 a month. Each month, Isaiah receives $250 from his parents an

inn [45]

Answer: 3

10x+475≥755

7 0

3 years ago

Read 2 more answers

The ordinate is twice the abscissa

zepelin [54]

We can construct the function:

$f(x)=2x$

Where $f(x)$ or $y$ is ordinate and $x$ is abscissa.

The function actually represents a line. Since it can have linear form.

$f(x)=2x+0$

Hope this helps.

r3t40

5 0

3 years ago

A four-digit number N leaves a remainder of 10 when divided by 21, a remainder of 11 when divided by 23, and a remainder of 12 w

aksik [14]

Answer:

Step-by-step explanation:

There may be a formal way to solve simultaneous Diophantine equations, but I don't know what it is. So, I addressed this by solving them 2 at a time.

First of all, we want to find a relation between integers p and q such that ...

21p+10 = 23q +11

21p -23q = 1 . . . . . . subtract 23q+10

Using the extended Euclidean algorithm or trial and error or astute observation, we find that (p, q) = (11, 10) is a solution to this equation. Then we can write p and q as ...

p = 23n+11
q = 21n+10

for some integer n.

Then our original numbers become ...

21p+10 = 21(23n+11)+10 = 483n +241 . . . . for some integer n

For this to be a 4-digit number, we must have n such that

1000 ≤ 483n +241 ≤ 9999

1.6 < n < 20.2

Modulo 25, the number (483n+241) is ...

8n +16

and we want that value to be 12:

(8n +16) mod 25 = 12

(8n +4) mod 25 = 0 . . . . . subtract 12

4(2n+1) mod 25 = 0 . . . . factor out 4

For this to be true, we must have 2n+1 be a multiple of 25. The only value of n that is in the required interval [2, 20] is n=12.

When n=12, 483n+241 = 6037. The sum of digits of 6037 is 16.

3 0

3 years ago

The product of a number and 12 is nomore than 15

allsm [11]

Answer:

As an equation, the product of a number and 12 is no more than 15 would be:

12n ≤ 15

N would represent the number.

≤ means less than or equal to, which is the same as "no more than".

5 0

3 years ago