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3 years ago

For the sequence an=3n-5 what is the value of a10

Mathematics

2 answers:

-BARSIC- [3]3 years ago

7 0

Answer: The required value of $a_{10}$ is 25.

Step-by-step explanation: We are given a sequence whose n-th term is given as follows :

$a_n=3n-5.$

We are to find the value of $a_{10}.$

To find the value of $a_{10}.$ , we need to put n = 10 in the given expression.

So, substituting n = 10 in the given expression, we get

$a_{10}=3\times 10-5=30-5=25.$

Thus, the required value of $a_{10}$ is 25.

liq [111]3 years ago

6 0

Answer: The value of $a_{10}=25$

Step-by-step explanation:

The given sequence : $a_n=3n-5$ .

To find the value of $a_10$ , need to substitute the value n=10 in the above equation , we will get

$a_{10}=3(10)-5$

Open Parenthesis ,

$a_{10}=3\times10-5$

Solve product ,

$\Rightarrow\ a_{10}=30-5$

Solve subtraction,

$\Rightarrow\ a_{10}=25$

Therefore , the value of $a_{10}=25$

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−(−49) = −49 true or false?

vova2212 [387]

I hope this helps you

false

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3 years ago

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Harman [31]

Answer:

See below for proof.

Step-by-step explanation:

Given:

$y=\left(x+\sqrt{1+x^2}\right)^m$

First derivative

$\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If $f(g(x))$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=f'(g(x))\:g'(x)$\\\end{minipage}}$

$\boxed{\begin{minipage}{5 cm}\underline{Differentiating $x^n$}\\\\If $y=x^n$, then $\dfrac{\text{d}y}{\text{d}x}=xn^{n-1}$\\\end{minipage}}$

$\begin{aligned} y_1=\dfrac{\text{d}y}{\text{d}x} & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{2x}{2\sqrt{1+x^2}} \right)\\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{x}{\sqrt{1+x^2}} \right) \\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(\dfrac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}} \right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(x+\sqrt{1+x^2}\right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m\end{aligned}$

Second derivative

$\boxed{\begin{minipage}{5.5 cm}\underline{Product Rule for Differentiation}\\\\If $y=uv$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=u\dfrac{\text{d}v}{\text{d}x}+v\dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}$

$\textsf{Let }u=\dfrac{m}{\sqrt{1+x^2}}$

$\implies \dfrac{\text{d}u}{\text{d}x}=-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}$

$\textsf{Let }v=\left(x+\sqrt{1+x^2}\right)^m$

$\implies \dfrac{\text{d}v}{\text{d}x}=\dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^m$

$\begin{aligned}y_2=\dfrac{\text{d}^2y}{\text{d}x^2}&=\dfrac{m}{\sqrt{1+x^2}}\cdot\dfrac{m}{\sqrt{1+x^2}}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}\\\\&=\dfrac{m^2}{1+x^2}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\\\\ &=\left(x+\sqrt{1+x^2}\right)^m\left(\dfrac{m^2}{1+x^2}-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\right)\\\\\end{aligned}$

$= \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\right)\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)$

Proof

$(x^2+1)y_2+xy_1-m^2y$

$= (x^2+1) \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m$

$= \left(x+\sqrt{1+x^2}\right)^m\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m$

$= \left(x+\sqrt{1+x^2}\right)^m\left[m^2-\dfrac{mx}{\sqrt{1+x^2}}+\dfrac{mx}{\sqrt{1+x^2}}-m^2\right]$

$= \left(x+\sqrt{1+x^2}\right)^m\left[0]$

$= 0$

8 0

2 years ago

What is the answer to this?

Pie

Answer:

the third one

Step-by-step explanation:

p n p is p

p n q is p

q n q is q

q n p is p

p: true

q: false

got it?

6 0

3 years ago

Please help ASAP Due very soon

MissTica

Answer:

y = -3/4 + 4

Step-by-step explanation:

use rise over run to find slope

6 0

3 years ago

HELP ASAP

Alex787 [66]

v(x)=(s/t)
= (3x - 6) / (-3x+6)
= [3(x-2)] / [-3(x-2)] --> 3 is factored out
= 1/-1 ---> common terms are cancelled out.
= -1 ---> This is the simplified formula.

To find the domain, we equate the denominator to 0.
-3x+6 = 0
3x = 6
x = 2
Domain: all values except 2.

w(x)=(t/s)(x)
= (-3x+6)x / (3x-6)
= [-3x(x-2)] / [3(x-2)] --> 3 is factored out
= -x --> The common terms are cancelled out. This is the simplified formula.

Solving for domain:
3x-6 = 0
3x = 6
x = 2
Domain: all values except 2.

4 0

3 years ago