MATHEMATICS HIGH SCHOOL

Name the subset(s) of real numbers to which each number belongs. Then order the numbers from least to greatest. 105−−−√,−4,43

Answers

Answer 1

Answer:

Answer:

(a)

$√(105) => Irrational$

$-4 => Integer$

$(4)/(3) => Rational$

(b)

$-4$ $(4)/(3)$ $√(105)$

Step-by-step explanation:

Given

$√(105)$

$-4$

$(4)/(3)$

Solving (a): The category of number the belong to

$√(105)$

Solve the square root

$√(105) = 10.246950766$

The result is irrational;

So:

$√(105) => Irrational$

$-4 => Integer$

$(4)/(3)$

This has an integer numerator and denominator;

So:

$(4)/(3) => Rational$

Solving (b): Order from least to greatest

i. $√(105) = 10.246950766$

ii. $-4$

iii. $(4)/(3) = 1.33333$

List out the corresponding numbers:

10.246950766; -4 and 1.333333

Reorder: from least to greatest:

-4; -1.3333333 and 10.246950766

Hence; The correct order is:

$-4$ $(4)/(3)$ $√(105)$

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Is 20,25,30 a right triangle?

I will mark the CORRECT answer Brainiest.

Answers

Answer:

Step-by-step explanation:

$20^2+25^2\n=400+625\n=1025$

However 1025 isn't a square number.

Should i restart my cumputer to see if my khan % will go up if i keep getting 100 on my work

Answers

Answer:

whats you mean

Step-by-step explanation:

Answer
I believe refreshing your page would be enough

A rectangular cartoon is 80 cm x60 cm x40 cm .then, how many packets of shops can each of 10 cm x 5cm x 4 cm be kept inside the cartoon

Answers

Answer:

Step-by-step explanation:

10 x 5 x 4

Multiply everything by 8

80 x 40 x 32

You can't go over that because 80 is at it's limit.

Explicación paso a paso:

10 x 5 x 4

Multiplica todo por 8

80 x 40 x 32

No se puede pasar de eso porque 80 está en su límite.

espero te sirva

Which long division problem can be used to prove the formula for factoring the difference of two perfect cubes?

Answers

Some of the possible options of the questions are;

A) $(a - b) | \overline {a^2 + a \cdot b + b^2}$

B) $(a + b) | \overline {a^2 - a \cdot b + b^2}$

C) $(a + b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$

D) $(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$

The difference of two perfect cubes has a binomial factor and a trinomial factor

The option that gives the long division problem that can be used to prove the difference of two perfect cubes is option D

D) $\underline {(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}}$

Reason:

The formula for factoring the difference of twoperfect cubes is presented as follows;

a³ - b³ = (a - b)·(a² + a·b + b²)

Given that a factor of the difference of two cubes is (a - b), and that we

have; (a³ + 0·a·b² + 0·a²·b - b³) = (a³ - b³), both of which are present in

option D, by long division of option D, we have;

${} \hspace {33} a^2 + a \cdot b + b^2\n(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a^2 \cdot b - b^3}\n{} \hspace {33} \underline{a^3 - a^2 \cdot b }\n{} \hspace {55} a^2 \cdot b + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3\n {} \hspace {55} \underline{a^2 \cdot b - a \cdot b^2}\n{} \hspace {89} a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3\n{} \hspace {89} \underline{a \cdot b^2 - b^3}\n{}\hspace {89} 0$

By the above long division, we have;

$(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$ = a² + a·b + b²

Which gives;

$(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$ = (a³ + 0·a·b² + 0·a·b² - b³)/(a - b)

We get;

(a³ + 0·a·b² + 0·a·b² - b³)/(a - b) = a² + a·b + b²

(a - b)·(a² + a·b + b²) = (a³ + 0·a·b² + 0·a·b² - b³) = (a³ - b³)

(a - b)·(a² + a·b + b²) = (a³ - b³)

(a³ - b³) = (a - b)·(a² + a·b + b²)

Therefore;

The long division problem that can be used to prove the formula for

factoring the difference of two perfect cubes is

$(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$ , which is option D

D) $(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$

Learn more here:

brainly.com/question/17022755

Answer:

The correct options, rearranged, are:

Options:

$A)(a^2+ab+b^2)/(a-b)\n\nB)(a^2-ab+b^2)/(a+b)\n\nC)(a^3+0a^2+0ab^2-b^3)/(a+b))\n\n D)(a^3+0a^2+0ab^2-b^3)/(a-b)$

And the asnwer is the last option (D).

Explanation:

You need to find which long division can be used to prove the formula for factoring the difference of two perfect cubes.

The difference of two perfect cubes may be represented by:

$a^3-b^3$

And it is, as a very well known special case:

$a^3-b^3=(a-b)(a^2+ab+b^2)$

Then, to prove, it you must divide the left side, $a^3-b^3$ , by the first factor of the right side, $a-b$

Note that, to preserve the places of each term, you can write:

$(a^3-b^3)=(a^3+0a^2+0ab^2-b^3)$

Then, you have:

$(a^3+0a^2+0ab^2-b^3)=(a-b)(a^2+ab+b^2)$

By the division property of equality, you can divide both sides by the same factor, which in this case will be the binomial, and you get:

$(a^3+0a^2+0ab^2-b^3)/(a-b)=(a^2+ab+b^2)$

That is the last option (D).

Rational numbers are _____ natural numbers. always sometimes never

Answers

They are sometimes natural numbers. For example, numbers like 2 and 3 can also be written as 2/1 and 3/1.

1/2 and 2/3 are rational numbers but they aren't natural numbers since they are not positive whole numbers.

So, not every rational number will be a natural number.

Answer:

sometimes

Step-by-step explanation:

Can anyone help me please?
What is
21 times x = 7