Tag: unchanging relations

Quizzes (2) Questions (72)

Questions Related to unchanging relations

Multiple choice maths unchanging relations algebra aid introduction to unknowns measures and relations

State True or False, if the following expression is polynomial in one variable.

$3\sqrt t+t\sqrt 2$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The expression $3\sqrt t+t\sqrt2$ contain the term $3\sqrt t$, here exponent of $t$ is $\dfrac12$, which is not a whole number.
Therefore, the given expression is not a polynomial in one variable.

Multiple choice maths unchanging relations algebra aid introduction to unknowns measures and relations
State True or False, if the following expression is polynomial in one variable (State reason for your answer):
$y+\dfrac {2}{y}$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The expression $y+\dfrac2y$ contain the term $\dfrac2y$, here

the exponent of $y$ is $-1$, which is not a whole number.
Therefore, the given expression is not a polynomial in one variable.

Multiple choice maths unchanging relations algebra aid introduction to unknowns measures and relations

State whether true/false:

The following expression is a polynomial in one variable:
$x^{10}+y^3+t^{50}$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Clearly, the given expression $x^{10}+y^3+t^{50}$ contains three variables $x,y\space and\space t$
Hence, the given expression is not a polynomial in one variable instead in three variables.

Multiple choice maths unchanging relations algebra aid introduction to unknowns measures and relations

If $\displaystyle A=\pi \left ( R^{2}-r^{2} \right )$, then $R$ is equal to

  1. $\displaystyle \sqrt{\frac{A-\pi r^{2}}{\pi }}$
  2. $\displaystyle \sqrt{\frac{A+\pi r^{2}}{\pi }}$
  3. $\displaystyle \sqrt{\frac{r^{2}\pi -A}{\pi }}$
  4. $\displaystyle \sqrt{\frac{r^{2}\pi -A}{r}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given, $A=\pi(R^2-r^2)$
Therefore, $A =$ $\displaystyle \pi R^{2}-\pi r^{2}$
$\Rightarrow  A+\pi r^{2}=\pi R^{2}$
$\displaystyle \Rightarrow R^{2}=\frac{A+\pi r^{2}}{\pi }$
$\displaystyle \Rightarrow $ $\displaystyle R=\sqrt{\frac{A+\pi r^{2}}{\pi }}$
Multiple choice maths unchanging relations algebra aid introduction to unknowns measures and relations

The sum of the reciprocals of $\displaystyle\frac{x+3}{x^2+1}$ and $\displaystyle\frac{x^2-9}{x^2+3}$ is

  1. $\displaystyle\frac{x^3+2x^2-x}{x^2-9}$
  2. $\displaystyle\frac{x^3-2x^2+x}{x^2-9}$

  3. 1

  4. 0
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Reciprocals will be $\frac { { { x }^{ 2 } }+1 }{ x+3 } $,$\frac { { x }^{ 2 }+3 }{ { x }^{ 2 }-9 } $
Their sum will be
$\frac { { { x }^{ 2 } }+1 }{ x+3 } +\frac { { x }^{ 2 }+3 }{ { x }^{ 2 }-9 } $
 $=\frac { \left( x-3 \right) \left( { x }^{ 2 }+1 \right) +{ x }^{ 2 }+3 }{ { x }^{ 2 }-9 } $
$=\frac { { x }^{ 3 }+x-3{ x }^{ 2 }-3+{ x }^{ 2 }+3 }{ { x }^{ 2 }-9 } $
 $=\frac { { x }^{ 3 }-2{ x }^{ 2 }+x }{ { x }^{ 2 }-9 } $

Multiple choice maths unchanging relations algebra aid introduction to unknowns measures and relations

If $\displaystyle x^{2}-3x+1=0$ then the value of $\displaystyle x-\frac{1}{x}$ is

  1. $\displaystyle \sqrt{5}$
  2. $\displaystyle \sqrt{3}$
  3. $\displaystyle \sqrt{2}$
  4. $\displaystyle \sqrt{6}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$x^{2}-3x+1$

$\therefore x^{2}+1=3x\Rightarrow \frac{x^{2}+1}{x}=\frac{3x}{x}\Rightarrow x+\frac{1}{x}=3$
$x^{2}+\frac{1}{x}^{2}=\left ( x+\frac{1}{x} \right )^{2}-2=(3)^{2}-=9-2=7$
We know 
$\left ( x-\frac{1}{x} \right )^{2}=x^{2}+\frac{1}{x}^{2}-2\Rightarrow 7-2=5$
$\therefore x-\frac{1}{x}=\sqrt{5}$

Multiple choice maths unchanging relations algebra aid introduction to unknowns measures and relations

If $\displaystyle x-\frac{1}{x}=3$; then the value of $\displaystyle \frac{3x^{2}-3}{x^{2}+2x-1}$ is

  1. 9/5

  2. 8/5

  3. 7/5

  4. 6/5
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given $x+\frac{1}{x}=3$ multiply by x both sides

Then $x^{2}-1=3x$
$\Rightarrow x^{2}-3x-1=0$
So $\frac{3x^{2}-3}{x^{2}+2x-1}=\frac{3(x^{2}-1)}{x^{2}-3x-1+5x}= \frac{3\times 3x}{0+5x}= \frac{9x}{5x}=\frac{9}{5}$

Multiple choice maths unchanging relations algebra aid introduction to unknowns measures and relations

If $\displaystyle x+\frac{a}{x}=b$ then the value of $\displaystyle \frac{x^{2}+bx+a}{bx^{2}-x^{3}}$ is

  1. $\displaystyle \frac{6b}{a}$
  2. $\displaystyle \frac{5b}{a}$
  3. $\displaystyle \frac{2b}{a}$
  4. $\displaystyle \frac{4b}{a}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given $x+\frac{a}{x}=b$ Multiply by x both sides

$x^{2}+a=bx$
$\Rightarrow x^{2}-bx+a=0$
So $\frac{x^{2}+bx+a}{bx^{2}-x^{3}}=\frac{x^{2}-bx+a+2bx}{-x(x^{2}-bx)}=\frac{0+2bx}{-x(-a)}=\frac{2bx}{ax}=\frac{2b}{a}$