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Ivan
3 years ago
5

Can someone find the mean of these numbers please? Thank you!

Mathematics
1 answer:
swat323 years ago
8 0
26.312. Add all of them up and divide by 10 since there are 10 numbers there


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A table<br> cloth is 82 inches<br> long. What is its length in<br> feet and inches?
Ratling [72]
The answer is 6 feet 10 inches
4 0
2 years ago
Please help its from the lines and angles chapter
Murrr4er [49]

This is the answer of this question

4 0
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If <img src="https://tex.z-dn.net/?f=%5Cmathrm%20%7By%20%3D%20%28x%20%2B%20%5Csqrt%7B1%2Bx%5E%7B2%7D%7D%29%5E%7Bm%7D%7D" id="Tex
Harman [31]

Answer:

See below for proof.

Step-by-step explanation:

<u>Given</u>:

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

<u>First derivative</u>

\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}}

<u />

<u />\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}}

<u />

\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}

<u>Second derivative</u>

<u />

\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)

<u>Proof</u>

  (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
1 year ago
What is the product of 2/3 * 3/8?
klio [65]

Answer:

0.25

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
The top of two poles of height's 20m and 14m are connected by the wire
Hatshy [7]

Answer:

The correct option is;

10 m

Step-by-step explanation:

The parameters given are;

Height of the first pole = 20 m

Height of the second pole = 14 m

The angle the wire connected to their top makes with the horizontal = 30°

The vertical height subtended by the inclined wire, h = The difference in height between the two poles = 20 - 14 = 6 m.

Let the horizontal distance between the two poles = D

Therefore;

The horizontal length of the wire = D

From trigonometric ratios, we have;

tan(30 ^{\circ}) = \dfrac{h}{D} = \dfrac{6}{D}

Which gives;

D = \dfrac{6}{tan(30 ^{\circ}) }  = 10.39 \ m\approx 10 \ m

The correct option is 10 m.

5 0
3 years ago
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