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3 years ago

What is the 22nd term of the arithmetic series where a1=8 and a9=56?

1 answer:

7 0

A 1 = 8, a 9 = 56
a 9 = a 1 + 8 d
56 = 8 + 8 d
8 d = 48
d = 6 ( common difference )
a 22 = a 1 + 21 d = 8 + 21 · 6 = 8 + 126 = 134

SO the answer is the 22nd term is 134

I really hope this helps!

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inysia [295]

$\large \bigstar \frak{ } \large\underline{\sf{Solution-}}$

Consider, LHS

$\begin{gathered}\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }} \\ \end{gathered} \\ \\ \text{So, using this identity, we get} \\ \\ \begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - ( {sec}^{2}\theta - {tan}^{2}\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

We know,

$\begin{gathered}\boxed{\sf{ \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: \: }} \\ \end{gathered} \\$

So, using this identity, we get

$\begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - (sec\theta + tan\theta )(sec\theta - tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}$

can be rewritten as

$\begin{gathered}\rm\:=\:\dfrac {(\sec \theta + tan\theta ) - (sec\theta + tan\theta )(sec\theta -tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac {(\sec \theta + tan\theta ) \: \cancel{(1 - sec\theta + tan\theta )}} { \cancel{ \tan \theta - \sec \theta + 1} } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:sec\theta + tan\theta \\\end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \\ \end{gathered} \\ \\ \\\begin{gathered}\rm \: = \:\dfrac{1 + sin\theta }{cos\theta } \\ \end{gathered}$

<h2>Hence,</h2>

$\begin{gathered} \\ \rm\implies \:\boxed{\sf{ \:\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \:\dfrac{1 + sin\theta }{cos\theta } \: \: }} \\ \\ \end{gathered}$

$\rule{190pt}{2pt}$

5 0

2 years ago

Is the set of integers closed under division? Explain why or why not.

MrMuchimi

The set of integers is not closed under the operation of division because when an integer is divided by another, the results are not limited to only integers i.e. zero and decimals are possibilities

<h3>How to determine the true statement?</h3>

Integers are numbers without decimal points

When integers are divided, the possible results are:

Zero
Decimal
Integers

The decimal and the zero possibilities imply that the set of integers is not closed under the operation of division

This is so because when an integer is divided by another, the results are not limited to only integers i.e. zero and decimals are possibilities

Read more about sets at:

brainly.com/question/1563195

#SPJ1

5 0

1 year ago

Create a quadratic expression in standard form

Georgia [21]

Answer:

ax²+bx+c

ax²+bxy+cy²

Step-by-step explanation:

A quadratic expression is one where the maximum degree of the variable or variables ( in case of more than one variable the sum of the degree of the variables in a single term considered) is 2.

For example:

A quadratic expression of a single variable is ax²+bx+c {Where a, b, and c are the arbitrary constants}

A quadratic expression with two variables is ax²+bxy+cy² {Where a, b, and c are the arbitrary constants}

And, a quadratic expression with three variables is ax² +by² +cz² +dxy +eyz +fzx {Where a, b, c, d, e, f are the arbitrary constants} (Answer)

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