If $z$ is a complex number such that $vertical line equals vertical line greater or equal than 2$ then the minimum value of $∣ z plus fraction numerator 1 over denominator 2 end fraction$ is is strictly

Therefore the correct option is choice 3

One of the values of $open parentheses cis invisible function application fraction numerator pi over denominator 6 end fraction close parentheses to the power of fraction numerator 1 over denominator 2 end fraction end exponent plus open parentheses cis invisible function application fraction numerator negative pi over denominator 6 end fraction close parentheses to the power of fraction numerator 11 over denominator 2 end fraction end exponent$

If the distance between the points $left parenthesis a c o s invisible function application theta comma a s i n invisible function application theta right parenthesis$ , $left parenthesis a c o s invisible function application ϕ comma a s i n invisible function application ϕ right parenthesis$ is $2 a$
then $theta equals$ 6
Assertion (A): If $2 s i n invisible function application fraction numerator theta over denominator 2 end fraction equals square root of 1 plus s i n invisible function application theta end root plus square root of 1 minus s i n invisible function application theta end root$ , then $fraction numerator theta over denominator 2 end fraction$ lies between
$2 n pi plus fraction numerator pi over denominator 4 end fraction text and end text 2 n pi plus fraction numerator 3 pi over denominator 4 end fraction left parenthesis n element of z right parenthesis$
Reason $left parenthesis R right parenthesis colon$ If $fraction numerator theta over denominator 2 end fraction element of open parentheses fraction numerator pi over denominator 4 end fraction comma fraction numerator 3 pi over denominator 4 end fraction close parentheses$ , then $s i n invisible function application fraction numerator theta over denominator 2 end fraction greater than 0$

If $c o s e c to the power of 6 end exponent invisible function application q minus c o t to the power of 6 end exponent invisible function application q equals a c o t to the power of 4 end exponent invisible function application q plus b c o t to the power of 2 end exponent invisible function application q plus c$ then a+b+c=

So here we have used the trigonometric functions and trigonometric formulas to solve this, the algebraic expressions were used to formulate it. Here the answer of a+b+c is 7.

If $c o s e c to the power of 6 end exponent invisible function application q minus c o t to the power of 6 end exponent invisible function application q equals a c o t to the power of 4 end exponent invisible function application q plus b c o t to the power of 2 end exponent invisible function application q plus c$ then a+b+c=

So here we have used the trigonometric functions and trigonometric formulas to solve this, the algebraic expressions were used to formulate it. Here the answer of a+b+c is 7.

If a,b,c are the sides of the triangle ABC such that $open parentheses 1 plus fraction numerator b minus c over denominator a end fraction close parentheses to the power of a end exponent open parentheses 1 plus fraction numerator c minus a over denominator b end fraction close parentheses to the power of b end exponent open parentheses 1 plus fraction numerator a minus b over denominator c end fraction close parentheses to the power of c end exponent greater or equal than 1$
Then the triangle $A B C$ must be

If $s i n invisible function application y equals x s i n invisible function application left parenthesis a plus y right parenthesis$ and $fraction numerator d y over denominator d x end fraction equals fraction numerator A over denominator 1 plus x to the power of 2 end exponent minus 2 x c o s invisible function application a end fraction$ then the value of $blank to the power of ´ end exponent A$ ' is

Therefore the correct option is choice 3

If $s i n invisible function application y equals x s i n invisible function application left parenthesis a plus y right parenthesis$ and $fraction numerator d y over denominator d x end fraction equals fraction numerator A over denominator 1 plus x to the power of 2 end exponent minus 2 x c o s invisible function application a end fraction$ then the value of $blank to the power of ´ end exponent A$ ' is

Therefore the correct option is choice 3

The maximum value of $open parentheses c o s invisible function application alpha subscript 1 end subscript close parentheses open parentheses c o s invisible function application alpha subscript 2 end subscript close parentheses horizontal ellipsis horizontal ellipsis. open parentheses c o s invisible function application alpha subscript n end subscript close parentheses$ under the restriction
$0 less or equal than alpha subscript 1 end subscript comma alpha subscript 2 end subscript horizontal ellipsis horizontal ellipsis horizontal ellipsis subscript n end subscript less or equal than fraction numerator pi over denominator 2 end fraction & c o t invisible function application alpha subscript 1 end subscript times c o t invisible function application alpha subscript 2 end subscript horizontal ellipsis horizontal ellipsis c o t invisible function application alpha subscript n end subscript equals 1$ is

The mean of five observations is 4 and their variance is 5.2. If three of these observations are 1, 2, and 6. Then the other two are

$text If end text x subscript 1 end subscript comma x subscript 2 end subscript comma x subscript 3 end subscript$ are three non zero real numbers such that $open parentheses x subscript 1 end subscript superscript 2 end superscript plus x subscript 2 end subscript superscript 2 end superscript close parentheses open parentheses x subscript 2 end subscript superscript 2 end superscript plus x subscript 3 end subscript superscript 2 end superscript close parentheses less or equal than open parentheses x subscript 1 end subscript x subscript 2 end subscript plus x subscript 2 end subscript x subscript 3 end subscript close parentheses to the power of 2 end exponent blank text then the G.M. of end text x subscript 1 end subscript comma x subscript 2 end subscript comma x subscript 3 end subscript text is end text$

When 10 is subtracted from all the observations, the mean is reduced to 60% of its value. If 5 is added to all the observations, then the mean will be

Hence, the new mean is 30.

When 10 is subtracted from all the observations, the mean is reduced to 60% of its value. If 5 is added to all the observations, then the mean will be

Hence, the new mean is 30.

If mean deviation through median is 15 and median is 450, then coefficient of mean deviation is

Hence, the coefficient of Mean Deviation is 1/30.

If mean deviation through median is 15 and median is 450, then coefficient of mean deviation is

Hence, the coefficient of Mean Deviation is 1/30.

The standard deviation of the data given by Variate (x) 0 1 2 3. . . . . n Frequency (f) $blank to the power of n end exponent C subscript 0 end subscript blank to the power of n end exponent C subscript 1 end subscript blank to the power of n end exponent C subscript 2 end subscript blank to the power of n end exponent C subscript 3 end subscript midline horizontal ellipsis times blank to the power of n end exponent C subscript n end subscript$

If the standard deviation of 0,1,2,3…..9 is K, then the standard deviation of 10,11,12,13…. 19 is

When 15 was subtracted from each of the seven observations the following number resulted : -3,0,-2,4,6,1,1. The mean of the distribution is

The standard deviation of a variable x is $sigma$ . The standard deviation of the variable $fraction numerator a x plus b over denominator c end fraction$ where a, b, c are constants is ...