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lawyer [7]
3 years ago
11

Is the relationship between the number of text in time a proportional relationship?

Mathematics
1 answer:
german3 years ago
3 0
Ratios are proportional if they represent the same relationship. One way to see if two ratios are proportional is to write them as fractions and then reduce them. If the reduced fractions are the same, your ratios are proportional.
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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
PLEASE ANSWER!!!!!! WILL MARK BRAINLIEST!!!!!!
docker41 [41]

Answer:

sinB= 3/5, tanA=4/3, cosB=4/5

Step-by-step explanation:

5 0
3 years ago
A can of soup has a radius of 6 inches and a height of 10 inches. How much soup can fill the can if it is filled to the top? Pla
denpristay [2]

A can of soup is a cylinder. We want to know how much soup can fill the can, which means that we are looking for the volume.

Formula for the volume of a cylinder: V = pi x r^2 x h

V = pi x 6^2 x 10

V = pi x 36 x 10

V = 360pi inches^3

Hope this helps! :)

6 0
3 years ago
Which inequality is represented by the graph?y &gt; -2/3x + 1y &lt; -2/3x + 1y &lt; -3/2x + 1y &gt; -3/2x + 1
Tanya [424]

Answer:

y < -\frac{3}{2}x+1

Step-by-step explanation:

Given:

From the graph shown,

The broken line means that the inequality sign will be either < or >.

The shaded region is to the left of the broken line. This means that y < f(x).

Now, in order to find equation of line, we need slope and y-intercept.

From the graph, the y intercept is at (0,1). So, y intercept is 1.

Also, if we consider two points (0,1)\textrm{ and }(2,-2) on the line, then slope is given as:

m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-2-1}{2-0}=\frac{-3}{2}=-\frac{3}{2}

Therefore, the equation of the line is given as:

y=mx+b\\y=-\frac{3}{2}x+1

But as the shaded region is to the left of the line,

∴ y is the correct answer.

7 0
3 years ago
Expand and simplify ( √2 + 3)to the power of 2
11111nata11111 [884]

Answer:

Step-by-step explanation:

(\sqrt{2}+3)^2=(\sqrt{2})^2 + 6\sqrt{2}+9 = 6\sqrt{2} +9+2=6\sqrt{2}+11

hope this helps

5 0
3 years ago
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