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Not correct! Here comes the correction.

49(x+1)squared minus 25  Factorize!

\(\large 49(x+1)^2-25\) 

\((7(x+1))^2-25=(7(x+1)+5)\times (7(x+1)-5)\)

\(=(7x+12)\times (7x+2)\)

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1.

\(y = t\sin t\\ x = t\cos t\)

2.

\(y = t\cos t\\ x = t\sin t\)

49(x+1)squared minus 25

\(\large 49(x+1)^2-25=7^2(x+1+5)(x+1-5)\) 

\(=7^2(x^2+x-5x+x+1-5+5x+5-25)\)

\(=7^2(x^2+2x-24) \)  

             \(x=1\pm\sqrt{1+24} \ \Rightarrow x_1=6 \ \wedge \ x_2=-4\)

\(\large =7\times 7 \times (x-6)\times (x+4) \)

!

49 (x+1)^2 = 25

(x+1)^2 = 25/49

x+1 = + - 5/7

x = 5/7 -1    or  -5/7 -1

x = -2/7 , -12/7

"Factorize by remarkable identities:49(x+1)squared minus 25"

\(49(x+1)^2-25\rightarrow(7x+7)^2-5^2\rightarrow(7x+7+5)(7x+7-5)\rightarrow(7x+12)(7x+2)\)

how to calculate parimetre of a circle

area A

radius r
diameter d
circumference C

\(C=d\pi\)

\(C=2\pi r\) 

\(C=\sqrt{4\pi A}\) 
\(A=\pi r^2\)

\(A=\frac{d^2 \pi}{4}\)

\(A=​ \frac{C^2}{4\pi}\)

\(r=\frac{C}{2\pi}\)

\(r=\sqrt{\frac{A}{\pi}}\)

\(r=​ \frac{d}{2}\)

\(d=​ \frac{C}{\pi}\) 

\(d=\sqrt{​ \frac{4A}{\pi}}\)

\(d=2r\) 

!

You need complex numbers to do this:

\((x-2-\sqrt7i)(x-2+\sqrt7i)=x^2-4x+11\)

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Plotting t*cos(t) on the x axis and t*sin(t) on the y axis does, indeed, generate a spiral.

There isn't much point in trying to express y in terms of x without the t as there are multiple values (an infinite number) of y for a given x (and vice-versa).

If you plot t*cos(t) on the y axis and t*sin(t) on the x axis you get a spiral that curves in the opposite direction.

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An architect build a model of a 220-foot tall building he is designning.

The model is 25 inches tall and 10 inches wide.

How wide is the actual building ?

\(\begin{array}{|rcll|} \hline \frac{10\ in.}{25\ in.} &=& 0.4 \\\\ \frac{\text{wide}}{220\ foot} &=& 0.4 \\ \text{wide} &=& 0.4\cdot 220\ foot \\ \text{wide} &=& 88\ foot \\ \hline \end{array}\)

The building is 88 foot wide

Do you have something between your ears, commonly known as a brain ?

If your graph plotter allows the use of moduli signs, you could experiment with the expressions

\(\displaystyle (|x|+x)/2 \text{ and } (|x|-x)/2\).

The first of these is equal to x when x is positive and zero when x is negative so it could replace the x in your first term.

The second is equal to zero when x is positive and x when x is negative, so would replace the x in your second term.

(That's a bit confusing, you should check the logic ! )

Tiggsy.

y=2x+1

\(y=2x+1\ [ \ is \ a \ function \ equation \ ]\)

  !

-4/5+(-3/4)(-3/4)(4)

\(-\frac{4}{5}+\frac{-3}{4}\times\frac{-3}{4}\times 4=-\frac{4}{5}+\frac{9}{4}=\frac{-16+45}{20}=\frac{29}{20}=1\frac{9}{20}\)

  !

what is -9 *1.6x10^-19

\(-9\times 1.6\times 10^{-19}=-1.44\times 10^{-20}\)

  !

see also same question:

I think I am supposed to use integration by substitution or integration by parts, but I'm not sure.

Can anyone help with this?

\(\int{ x^3\cdot\sqrt{x^2-1}\ dx } \)

short way:

Substitution

\(\begin{array}{|rcll|} \hline u &=& \sqrt{x^2-1}=(x^2-1)^{\frac12} \quad & \text{or} \quad x = \sqrt{u^2+1} \\ du &=& \frac12 \cdot (x^2-1)^{-\frac12} \cdot 2x\ dx \\ &=& \frac{x}{\sqrt{x^2-1}}\ dx \quad & \text{or} \quad dx = \frac{\sqrt{x^2-1}}{x}\ du\\ \hline \end{array} \begin{array}{|rcll|} \hline && \int{ x^3\cdot\sqrt{x^2-1}\ dx } \\ &=& \int{ ( \sqrt{u^2+1})^3\cdot u \frac{\sqrt{x^2-1}}{x}\ du } \\ &=& \int{ ( \sqrt{u^2+1})^3\cdot u \frac{u}{\sqrt{u^2+1}}\ du } \\ &=& \int{ ( \sqrt{u^2+1})^2\cdot u^2\ du }\\ &=& \int{ (u^2+1)\cdot u^2\ du }\\ &=& \int{ (u^4+u^2)\ du }\\ &=& \frac{u^3}{3} + \frac{u^5}{5} + c \\ &=& \frac{u^3}{3}\cdot \frac55 + \frac{u^5}{5}\cdot \frac33 + c \\ &=& \frac{u^3}{15}\cdot \left( 5+3u^2 \right)+ c \\ \hline \end{array} \)

Back substitution

\(\begin{array}{|rcll|} \hline u &=& \sqrt{x^2 -1} \\ \hline \end{array} \begin{array}{|rcll|} \hline && \mathbf{\int{ x^3\cdot\sqrt{x^2-1}\ dx } }\\ &=& \frac{u^3}{15}\cdot \left( 5+3u^2 \right)+ c \\ &=& \frac{(\sqrt{x^2 -1})^3}{15}\cdot \left[ 5+3(\sqrt{x^2 -1})^2 \right]+ c \\ &=& \frac{(x^2 -1)^{\frac32}}{15}\cdot \left[ 5+3(x^2 -1) \right]+ c \\ &=& \frac{(x^2 -1)^{\frac32}}{15}\cdot ( 5+3x^2-3 )+ c \\ &\mathbf{=}& \mathbf{ \dfrac{(x^2 -1)^{\frac32}}{15}\cdot ( 3x^2+2 )+ c }\\ \hline \end{array} \)

\(0.\overline{33}\)

It won't round to 6.07...it will either round to 6.1  [ if we're rounding to the nearest tenth]  or to 6.08 [if we're rounding to the nearest hundredth]  or to  6 [ if it's to the nearest integer ]

I think I am supposed to use integration by substitution or integration by parts, but I'm not sure. Can anyone help with this?

\(\int{x}^{3}\sqrt{{x}^{2}-1}dx\)

We apply twice substitution

We need the following formulae:

\(\begin{array}{|rcll|} \hline \cosh^2(z) - \sinh^2(z) &=& 1 \\ \sinh^2(z) &=& \cosh^2(z) -1 \\ \sinh(z) &=& \sqrt{\cosh^2(z) -1} \\\\ \cosh^2(z) - \sinh^2(z) &=& 1 \\ \cosh^2(z) &=& 1+\sinh^2(z) \\\\ \frac{d}{dz}\sinh(z) &=& \cosh(z) \\ \frac{d}{dz}\cosh(z) &=& \sinh(z) \\ \hline \end{array} \)

1. Substitution

\(\begin{array}{|rcll|} \hline x &=& \cosh(z)\\ dx &=& \sinh(z)\ dz \\ \hline \end{array} \begin{array}{|rcll|} \hline && \int{ x^3\cdot\sqrt{x^2-1}\ dx } \\ &=& \int{ \cosh^3(z)\cdot \sqrt{\cosh^2(z)-1} \cdot \sinh(z)\ dz } \\ &=& \int{ \cosh^3(z)\cdot \sinh(z) \cdot \sinh(z)\ dz } \\ &=& \int{ \cosh^3(z)\cdot \sinh^2(z)\ dz } \\ \hline \end{array} \)

2. Substitution

\(\begin{array}{|rcll|} \hline u &=& \sinh(z)\\ du &=& \cosh(z)\ dz \\ \hline \end{array} \begin{array}{|rcll|} \hline && \int{ \cosh^3(z)\cdot \sinh^2(z)\ dz } \\ &=& \int{ \cosh^2(z)\cdot \sinh^2(z)\cdot \cosh(z)\ dz \\ } \\ &=& \int{ [1+\sinh^2(z)]\cdot \sinh^2(z)\cdot \cosh(z)\ dz \\ } \\ &=& \int{ (1+u^2)\cdot u^2 \ du } \\ &=& \int{ (u^2+u^4)\ du } \\ &=& \frac{u^3}{3} + \frac{u^5}{5} + c \\ &=& \frac{u^3}{3} + \frac{u^5}{5} + c \\ &=& \frac{u^3}{3}\cdot \frac55 + \frac{u^5}{5}\cdot \frac33 + c \\ &=& \frac{u^3}{15}\cdot \left( 5+3u^2 \right)+ c \\ \hline \end{array} \)

3. Back substitution

\(\begin{array}{|rcll|} \hline u &=& \sinh(z) \\ &=& \sqrt{\cosh^2(z) -1} \\ &=& \sqrt{x^2 -1} \\ \hline \end{array} \begin{array}{|rcll|} \hline && \mathbf{\int{ x^3\cdot\sqrt{x^2-1}\ dx } }\\ &=& \frac{u^3}{15}\cdot \left( 5+3u^2 \right)+ c \\ &=& \frac{(\sqrt{x^2 -1})^3}{15}\cdot \left[ 5+3(\sqrt{x^2 -1})^2 \right]+ c \\ &=& \frac{(x^2 -1)^{\frac32}}{15}\cdot \left[ 5+3(x^2 -1) \right]+ c \\ &=& \frac{(x^2 -1)^{\frac32}}{15}\cdot ( 5+3x^2-3 )+ c \\ &\mathbf{=}& \mathbf{ \dfrac{(x^2 -1)^{\frac32}}{15}\cdot ( 3x^2+2 )+ c }\\ \hline \end{array} \)

x = sec θ     dx  =  sec θ*tan θ dθ

∫  sec^3( θ )  √[ sec^2θ  - 1 ] secθ*tanθ dθ  =

∫ sec^3( θ ) √ [ tan^2θ ]  secθ*tanθ dθ  =

∫ sec^3( θ ) √ [ tanθ ]  secθ*tanθ dθ  =

∫ sec^4 θ *  tan^2θ   dθ

∫ [sec^2 θ] * [sec^2 θ] * tan^2θ   dθ  =

∫ tan^2θ * [ tan^2θ + 1] [sec^2 θ]   dθ  =

∫ [  tan^4θ  +  tan^2θ ]  [sec^2 θ]   dθ  =

∫ [ tanθ ]^4 *  [sec^2 θ]   dθ    +  ∫   [ tanθ ]^2 *  [sec^2 θ]  dθ  =

(1/5) [ tanθ ]^5    +  (1/3)  [tanθ ]^3   + C

Now  x  = sec θ     ...so....   x^2  = sec^2θ  =  1 +  [ tanθ]^2  → 

x^2  - 1  =   [tanθ]^2     ...so......

[x^2  - 1]^(5/2)  = [ [tanθ]^2 ] ^(5/2)  = [ tanθ ]^5

And

[x^2  - 1]^(3/2)  = [ [tanθ]^2 ] ^(3/2)  = [ tanθ ]^3

So we have

∫  x^3 √[ x^2  - 1 ]  dx  =

(1/5)[x^2  - 1]^(5/2)  + (1/3)[x^2  - 1]^(3/2)   + C

Proof :

http://www.wolframalpha.com/input/?i=derivative++(1%2F5)%5Bx%5E2++-+1%5D%5E(5%2F2)++%2B+(1%2F3)%5Bx%5E2++-+1%5D%5E(3%2F2)+++%2B+C

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