X + y is rational, must x and y each be rational

2 answers:

7 0

The rationals are closed under addition, so if x,y are rational then x+y must be rational since Q is a group under addition.

timofeeve [1]3 years ago

3 0

Yes because x and y must be integers

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R = {10, 15, 20}

Umnica [9.8K]

Answer:

R n S = {20}

Step-by-step explanation:

n in sets means same elements found in the sets which are being compared

5 0

3 years ago

Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants. 37 x2 − 6x

Dvinal [7]

Answer:

$\dfrac{A}{x}+\dfrac{B}{x-6}$

Step-by-step explanation:

Given the function $\dfrac{37}{x(x-6)}$ , to write the form of its partial fraction on decomposition, we will separate the two functions separated by an addition sign. The numerator of each function will be constants A and b and the denominator will be the individual factors of each function at the denominator. The partial fraction of the rational function is as shown below.

$= \dfrac{37}{x(x-6)}\\\\= \dfrac{A}{x}+\dfrac{B}{x-6}$

<em>Since we are not to solve for the constants, hence the partial fraction is </em> $\dfrac{A}{x}+\dfrac{B}{x-6}$

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3 years ago

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riadik2000 [5.3K]

Answer:

6 buses and 3 vans

Step-by-step explanation:

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3 years ago

HELP! Find the value of sin 0 if tan 0 = 4; 180 < 0< 270

BabaBlast [244]

Hi there! Use the following identities below to help with your problem.

$\large \boxed{sin \theta = tan \theta cos \theta} \\ \large \boxed{tan^{2} \theta + 1 = {sec}^{2} \theta}$

What we know is our tangent value. We are going to use the tan²θ+1 = sec²θ to find the value of cosθ. Substitute tanθ = 4 in the second identity.

$\large{ {4}^{2} + 1 = {sec}^{2} \theta } \\ \large{16 + 1 = {sec}^{2} \theta } \\ \large{ {sec}^{2} \theta = 17}$

As we know, sec²θ = 1/cos²θ.

$\large \boxed{sec \theta = \frac{1}{cos \theta} } \\ \large \boxed{ {sec}^{2} \theta = \frac{1}{ {cos}^{2} \theta} }$

And thus,

$\large{ {cos}^{2} \theta = \frac{1}{17}} \\ \large{cos \theta = \frac{ \sqrt{1} }{ \sqrt{17} } } \\ \large{cos \theta = \frac{1}{ \sqrt{17} } \longrightarrow \frac{ \sqrt{17} }{17} }$