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3 years ago

Let cos x =1√10 with x in QIV and find: cot2x

1 answer:

3 0

Cos x = 1/√10
sin x = √( 1 - 1 /10 ) = √9/10 = - 3/√10 ( because x is in Q IV )
cot 2 x = cos 2 x/sin 2x = ( cos²x - sin²x)/ (2 sinx cosx ) =
= ( 1/10 - 3/10 )/(2 * (-3/√10)*1/√10)= (- 8/10 ) / (-6/10 ) = 8/6 = 4/3

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Write a rule for the nth term of the arithmetic sequence:<br><br> a11= 50, d = 7

ivolga24 [154]

Answer:

Required rule for $n^{th}$ is $a_{n}=7n-27$ .

Step-by-step explanation:

Given that,

$a_{11} = 50,\ \ d=7$

From the question: we have to write the $n^{th}$ term of Arithmetic sequence.

Arithmetic Sequence or Arithmetic progression (A.P) : It is a sequence which possess that difference between of two successive sequence is always constant.

$a_{1} ,a_{2},a_{3},a_{4}.....................a_{n-1},a_{n}$

where, $a_{1}$ is the first term of A.P

$d$ is the common difference.

$a_{n}$ is the last term or general term.

The above sequence to be in A.P then their common difference should be equal.

$d=a_{2}-a_{1} =a_{3}-a_{2}=a_{4} -a_{3} ..........................a_{n}-a_{n-1}$

Now, Formula of General Term is $a_{n}=a+(n-1)d$

So, $a_{11}= a+(11-1)d\\ a_{11} = a+10d$

Substituting the value of $a_{11} = 50,\ \ d=7$ we get,

$50=a+10\times7\\50=a+70\\a=-20$

Then General term ( $a_{n}$ ) of given data is

$a_{n}=-20+(n-1)7\\a_{n}=-20+7n-7\\a_{n}=7n-27$

Therefore, Required rule for $n^{th}$ is $a_{n}=7n-27$ .

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3 years ago

can i create this painting by a 200 feet paper roll yes or no? or do i need more to create it its for a project on learning pixe

ioda

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