All Answers 356000 Answers

6x−y+3z = 28

5x−2y+2z = 28

−3x−y+4z = -17

cosec(10)+cosec(50)-cosec(70)=6

This is true but I would like to see how a mathematician can prove it as at present it has me baffled.

We are given a regular heptagon of side length 1. Let S be the set of points that are within a distance of 1 from some point on or inside the heptagon, but not including the heptagon itself. Find the area of S

\(\begin{array}{|rcll|} \hline A_1 &=& 1\cdot 1 \\ &=& 1 \\\\ A_2 &=& \pi \cdot 1^2 \cdot \frac{ \frac{360^{\circ}}{7} } {360^{\circ}} \\ &=& \pi \cdot \frac{1}{7} \\\\ S &=& 7\cdot A_1 + 7 \cdot A_2 \\ &=& 7\cdot 1 + 7 \cdot \pi \cdot \frac{1}{7} \\ &=& 7+\pi \\ &=& 10.1415926536 \\ \hline \end{array}\)

There are seven rectangular sections, each of area 1, and seven curved sections each of which form one seventh of a sector of a unit circle.  Hence the area of S is 7 + 7*pi*1²/7 or S = 7 + pi

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evaluate cosec(10)+cosec(50)-cosec(70)

=6

((a - 1)^4 - (b + 1)^4)/((a - 1)^3 - (b + 1)^3)

=(a^3 + a^2 b - 2 a^2 + a b^2 + 2 a + b^3 + 2 b^2 + 2 b) / (a^2 + a b - a + b^2 + b + 1)

((a−1)4−(b+1)4)/((a−1)3−(b+1)3)

\(\large \frac{(a-1)^4-(b+1)^4}{(a-1)^3-(b+1)^3}\color{black}=\frac{[(a-1)^2+(b+1)^2]\times [(a-1)^2-(b+1)^2]}{a^3-3a^2+3a-b^3-3b^2-3b-2}\)

\(\large= \frac{(a^2-2a+1+b^2+2b+1)\times(a^2-2a+1-b^2-2b-1)}{a^3-3a^2+3a-b^3-3b^2-3b-2}\)

\(\large =\frac{a^4 - 4 a^3 + 6 a^2 - 4 a - b^4 - 4 b^3 - 6 b^2 - 4 b}{a^3-3a^2+3a-b^3-3b^2-3b-2}\)

\( Whoever \ can \ do \ it, \ makes \ something \ of \ it.\)

  !

Suppose that $216 is invested 2.33% interest compounded quarterly. How much is in the account after 6 years

FV = PV [ 1 + R]^N

FV =216 x [ 1 + 0.0233/4]^(6*4)

FV =216 x 1.005825^24

FV =216 x 1.1495774...

FV =$248.31

(?)^2 = 9^4

\((x)^2 = 9^4\\ (x)^2 = (9^2)^2\\ (x)^2 = (81)^2\\ x=\pm81 \)

(?)^2 = 9^4

x^2 = 6,561 take the square root of both sides

x = 81

e^17.3 * x / (x + 237.3) = 0.013 * (x + 273.2)

\(\frac{e^{17.3} \times x }{ (x + 237.3) }= 0.013 (x + 273.2)\\ e^{17.3} \times x = 0.013 (x + 273.2)(x + 237.3) \\\)

Just expand it all out, take it all to one side ie make =0

then solve using the quadratic formula :)

Answers to questions 2 to 5:

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Grad.   =   subtract the y coordinates in some order  /   subtract the x coordinates in same order

So  we have

[ 32 - 2 ] / [ 8 - 2]   =   30 / 6    =   5

Find the equation for the line passing through the point (-4,-1) and perpendicular to the line whose equation is -1 x + 2y + 4=0.

Let's find the slope of the given line

-1 x + 2y + 4=0      add 1x to both sides   subtract 4 from both sides

2y  =  1x - 4          divide by 2

y = (1/2)x -  2       the slope  = 1/2

And the slope of a perp. line will be the negative reciprocal of this  =  -2 / 1   = - 2

So  we have

y =  m ( x - x1) + y1         where m is the slope = - 2  and  (x1, y1)  = (-4,-1)

So we have

y  = -2 (x - -4)  + - 1

y = -2 (x + 4) - 1

y = -2x - 8 - 1

 y = -2x - 9

can anybody teach me thanks

Can anybody teach me this question please. Just tell me as fast as you can.Thx

This'll help later in the future ---> https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=finding+the+side+length+of+a+rectangle+given+its+perimeter+or+area+calculator&*

Length = 7.14
Width= 4.04

216 x 0.0233 = 5.0328

5.0328 x 4 (4 times a year ) = 20.1312

20.1312 x 6 (6 years) = 120.7872

Answer = $120.79

simplify tan[ 2*arccos(x/3) ]

\(\begin{array}{|rcll|} \hline \cos(\varphi) &=& \frac{x}{3} \\ \varphi &=& \arccos\left(\frac{x}{3} \right) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \tan(2 \varphi ) &=& \frac{\sin(2 \varphi)}{\cos(2 \varphi)} \quad & | \quad \sin(2 \varphi) = 2 \sin(\varphi)\cos(\varphi) \quad \cos(2\ \varphi) = 2 \cos^2(\varphi)-1 \\ \tan(2 \varphi ) &=& \frac{2 \sin(\varphi)\cos(\varphi)}{2 \cos^2(\varphi)-1} \quad & | \quad \cos(\varphi) = \frac{x}{3} \\ \tan(2 \varphi ) &=& \frac{2 \sin(\varphi)\frac{x}{3}}{2 (\frac{x}{3})^2-1} \quad & | \quad \sin(\varphi) = \sin(\arccos\left(\frac{x}{3} \right)) \\ \tan(2 \varphi ) &=& \frac{2 \sin(\arccos\left(\frac{x}{3} \right))\frac{x}{3}}{2 (\frac{x}{3})^2-1} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \tan(2 \varphi ) &=& \frac{2 \sin(\arccos\left(\frac{x}{3} \right))\frac{x}{3}}{2 (\frac{x}{3})^2-1} \quad & | \quad \sin(\arccos\left(\frac{x}{3} \right)) = \sqrt{1-\left(\dfrac{x}{3}\right)^2} \\ \tan(2 \varphi ) &=& \frac{2 \sqrt{1-\left(\frac{x}{3}\right)^2}\frac{x}{3}}{2 (\frac{x}{3})^2-1} \\ \tan(2 \varphi ) &=& \frac{2 \sqrt{\frac{9-x^2}{9}}\frac{x}{3}}{ \frac{2x^2-9}{9}} \\ \tan(2 \varphi ) &=& \frac{2 \frac{\sqrt{9-x^2}}{3}\frac{x}{3}}{ \frac{2x^2-9}{9}} \\ \tan(2 \varphi ) &=& \frac{2 \frac{\sqrt{9-x^2}}{9}x}{ \frac{2x^2-9}{9}} \\ \tan(2 \varphi ) &=& \frac{2 \sqrt{9-x^2} x}{ 2x^2-9 } \quad & | \quad \sqrt{9-x^2} = \sqrt{3-x}\cdot\sqrt{3+x}\\ \tan(2 \varphi ) &=& \frac{2 \sqrt{3-x}\cdot x \cdot \sqrt{3+x}}{ 2x^2-9 } \\ \mathbf{\tan(2 \arccos\left(\frac{x}{3} \right) ) }& \mathbf{=} & \mathbf{\frac{2 \sqrt{3-x}\cdot x \cdot \sqrt{3+x}}{ 2x^2-9 } } \\ \hline \end{array}\)

let f(x)=(3x2+2x+14)/(1-5x2). The horizontal asymptote is given by y=?

round to 1 decimal place

\(f(x)=\frac{3x^2+2x+14}{1-5x^2}\\ \text{The horizontal asymptotes is }\\ f(x)=\displaystyle\lim_{x\rightarrow \pm \infty}\frac{3x^2+2x+14}{1-5x^2}\\ f(x)=\;-\;\displaystyle\lim_{x\rightarrow \pm \infty}\frac{3x^2+2x+14}{5x^2-1}\\ \text{Now the numerator and the denominator tend to infinity}\\ \text{so use l'Hopital's theorum}\\ f(x)=\;-\;\displaystyle\lim_{x\rightarrow \pm \infty}\frac{6x+2}{10x}\\ f(x)=\;-\;\displaystyle\lim_{x\rightarrow \pm \infty}(\frac{6}{10}+\frac{2}{10x})\\ f(x)=\;-0.6 \)

You can also do this using algebraic division.  It is almost as easy. :)

(A x 15%) + A =27,850

.15A  + A   =  27850

A ( .15 + 1)  =  27850   divide both sides by (.15 + 1)

A  =    27850 / ( .15 + 1)  ≈  24217.39

tan (-3pi/ 4)   = 1   so we have

tan (-3pi/4)  =  y/ x  =  y / -1

1   =   y / -1       multiply both sides by -1

-1   = y

So....the point  is   (-1, -1)

It's 20 dont listen to them when they say it's 19 thats the wrong answer youll fail ur test

Mmm ok, here is some Turkish coffee,

It should be strong enough!

Assuming 365 days in a year

1,000,000 / 365   ≈  2739.7 years