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2 years ago

Is -1 ⅐ a natural, whole, integer, rational or irrational number?

Mathematics

1 answer:

lana66690 [7]2 years ago

5 0

Answer:

Rational

Step-by-step explanation:

okay so a natural number is anything like 1,2,3,4...etc.

a whole number is like a natural number but includes 0's

an integer is numbers that can have both positive and negative numbers

but keep in mind all of these are whole numbers or can be made into a whole number (not including natural numbers)

this leaves...

Rational

now here is a word of advice every mixed number and fraction is rational because its already in a solved and complete form but a decimal that could be solved like 0.77777 or 0.345 with a line over it or 64 squared

Irrational

a number that has no end and cant be solved like 0.86866968..... or 38 squared

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andreyandreev [35.5K]

First expression: (-7)/(-4)

Divide. Note that two negative numbers divided will result in a positive answer

(-7)/(-4) = 7/4 = 1.75

First expression: Greater than 1.

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Second expression: -(3/2)

Simplify: 3/2 = 1.5

-(1.5) = -1.5

Second expression: Less than -1

--------------------------------------------------------------------------------------------------------------

Third expression: (-8/5) x (-5/8)

Note that two negatives = one positive answer when multiplying

8/5 x 5/8 = 40/40 = 1

Third expression: Neither

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Fourth Expression: (-5)/(-3)

Divide: (-5)/(-3) = 5/3 = ~1.67

Fourth Expression: Greater than 1

----------------------------------------------------------------------------------------------------------------

Fifth Expression: (-9)/6

Divide: (-9)/6 = -1.5

Fifth expression: Less than -1

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hope this helps

5 0

3 years ago

(3/4)to the power of 3 in form of p/q

nydimaria [60]

<h3><u>Solution</u><u>:</u><u>-</u></h3>

(3/4)to the power of 3 in form of p/q

$\pmb{(\dfrac{3}{4})^{^{^{3}}}=\dfrac{p}{q} }$

$\pmb{(\dfrac{27}{64})=\dfrac{p}{q} }$

8 0

3 years ago

The factorization of 8x3 -125 is (2x-5)(jx2 +kx+25)

lbvjy [14]

$\bf ~\hspace{10em}\textit{difference and sum of cubes} \\\\ a^3+b^3 = (a+b)(a^2-ab+b^2) ~\hfill a^3-b^3 = (a-b)(a^2+ab+b^2) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 8x^3-125~~ \begin{cases} 8=2\cdot 2\cdot 2\\ \qquad 2^3\\ 125=5\cdot 5\cdot 5\\ \qquad 5^3 \end{cases}\implies 2^3x^3-5^3\implies (2x)^3-5^3 \\[2em] [2x-5][(2x)^2+(2x)(5)+5^2]\implies (2x-5)(\stackrel{\stackrel{j}{\downarrow }}{4}x^2+\stackrel{\stackrel{k}{\downarrow }}{10}x+25)$