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3 years ago

Let a and b be rational numbers. Then by definition

Mathematics

2 answers:

balu736 [363]3 years ago

7 0

Answer:

the product of two rational numbers is a rational number!

Step-by-step explanation:

ohaa [14]3 years ago

6 0

Answer:

The product of two rational numbers is a rational number

Step-by-step explanation:

I'll quickly recap the proof: a rational number is, by definition, the ratio between two integers. So, there exists four integers m,n,p,q such that

$a=\dfrac{m}{n},\quad b=\dfrac{p}{q}$

If we multiply the fractions, we have

$ab = \dfrac{mp}{nq}$

Now, mp and nq are multiplication of integers, and thus they are integers themselves. So, ab is also a ratio between integer, and thus rational.

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HELP PLEASE <br><br> how do i put this into a piecewise function?

Nonamiya [84]

Answer:

Step-by-step explanation:

2 functions start by making a domain

Function 1: -3≤x<1

Function 2: -1≤x≤1

Now come up with equations

Function 1= x+3

Function 2= 5

Now put it all together

f(x)= x+3 for -3≤x<1 and 5 for -1≤x≤1

8 0

3 years ago

Which statement is true?

love history [14]

<h2>Hello!</h2>

The answer is:

The second option,

$(\sqrt[m]{x^{a} } )^{b}=\sqrt[m]{x^{ab} }$

Discarding each given option in order to find the correct one, we have:

<h2>First option,</h2>

$\sqrt[m]{x}\sqrt[m]{y}=\sqrt[2m]{xy}$

The statement is false, the correct form of the statement (according to the property of roots) is:

$\sqrt[m]{x}\sqrt[m]{y}=\sqrt[m]{xy}$

<h2>Second option,</h2>

$(\sqrt[m]{x^{a} } )^{b}=\sqrt[m]{x^{ab} }$

The statement is true, we can prove it by using the following properties of exponents:

$(a^{b})^{c}=a^{bc}$

$\sqrt[n]{x^{m} }=x^{\frac{m}{n} }$

We are given the expression:

$(\sqrt[m]{x^{a} } )^{b}$

So, applying the properties, we have:

$(\sqrt[m]{x^{a} } )^{b}=(x^{\frac{a}{m}})^{b}=x^{\frac{ab}{m}}\\\\x^{\frac{ab}{m}}=\sqrt[m]{x^{ab} }$

Hence,

$(\sqrt[m]{x^{a} } )^{b}=\sqrt[m]{x^{ab} }$

<h2>Third option,</h2>

$a\sqrt[n]{x}+b\sqrt[n]{x}=ab\sqrt[n]{x}$

The statement is false, the correct form of the statement (according to the property of roots) is:

$a\sqrt[n]{x}+b\sqrt[n]{x}=(a+b)\sqrt[n]{x}$

<h2>Fourth option,</h2>

$\frac{\sqrt[m]{x} }{\sqrt[m]{y}}=m\sqrt{xy}$

The statement is false, the correct form of the statement (according to the property of roots) is:

$\frac{\sqrt[m]{x} }{\sqrt[m]{y}}=\sqrt[m]{\frac{x}{y} }$

Hence, the answer is, the statement that is true is the second statement:

$(\sqrt[m]{x^{a} } )^{b}=\sqrt[m]{x^{ab} }$

Have a nice day!

6 0

3 years ago

Those are the right answer or i don’t think so i really don’t know i. really need help

Whitepunk [10]

Answer:

Triangle is isosceles triangle X = 9

Step-by-step explanation:

5 0

3 years ago

Complete the proof by providing the missing statement and reasons

Lisa [10]

Answer:

In triangle SHD and triangle STD.

$\overline{SD} \perp \overline{HT}$ [Side]

Since, a line is said to be perpendicular to another line if the two lines intersect at a right angle.

⇒ $\angle SDH = \angle SDT = 90^{\circ}$

$\overline{SH} \cong \overline{ST}$ [leg] [Given]

Reflexive property states that the value is equal to itself.

$\overline{SD} \cong \overline{SD}$ [Leg] [Reflexive property]

HL(Hypotenuse-leg) theorem states that any two right triangles that have a congruent hypotenuse and a corresponding congruent leg are the congruent triangles.

then, by HL theorem;

$\triangle SHD \cong \triangle STD$ Proved!

6 0

3 years ago

Estimate and solve 3,836 ÷ 63 = ________. (2 points)<br> 60 r 9<br> 58 r 56<br> 60 r 56<br> 58 r 9

7nadin3 [17]

Answer:

60 r 56

Step-by-step explanation:

6 0

3 years ago