All Answers 356000 Answers

9x - 28 = (6x + 2) 

3x = 30 

x = 10

RSV = 9x - 28 → 62º

Find the range of the list:
(-8, -9, -5, -9, -7, -5)

The range of a list is given by:
(largest element) - (smallest element)

(-8, -9, -5, -9, -7, -5) We see that the largest element is -5:
-5 - (smallest element)

(-8, -9, -5, -9, -7, -5) We see that the smallest element is -9:
-(-9) - 5

-(-9) - 5 = 4:
Answer: | 4

Hi Guest,

You are lucky that I saw this continued question.  It would be a good idea for you to become a member because then you would be able to send me a private message with a link to this thread and ask me to come back to it again.

PLUS I could be much more sure that you see any late answer that may be added.  :)

Yes you can do algebraic division with more than one variable, I have seen other mathematicians use it and I can use it on this one but I am not sure how often it is useful....

Doing Algebraic division with LaTex is difficult.

I have done it before with great sucess but it was extremely time consuming. ://

I'll give it a go.

This is different from how I did it before, This is quicker but not quite as nice.

\(\begin{array}{llllll}\\ &&9x^2&\color{Maroon}{+4y^2}&\color{NavyBlue}{+6xy}&&\\\hline 3x-2y|&&27x^3&-8y^3\\ &&27x^3&&-18yx^2\\\hline &&&-8y^3&\color{NavyBlue}{+18yx^2}\\ &&&&+18yx^2&-12xy^2\\\hline &&&-8y^3&&\color{Maroon}{+12xy^2}\\ &&&-8y^3&&+12xy^2\\\hline &&&+0&&+0\\\hline \end{array}\)

Heureka or maybe Max

Could you please show me how to make the horizontal lines the correct length ??  

Solve for x:
0.013 (x + 237.3) (x + 273.2) = 3.26058×10^7 x

Expand out terms of the left hand side:
0.013 x^2 + 6.6365 x + 842.795 = 3.26058×10^7 x

Subtract 3.26058×10^7 x from both sides:
0.013 x^2 - 3.26058×10^7 x + 842.795 = 0

x = (3.26058×10^7 ± sqrt((-3.26058×10^7)^2 - 4×0.013×842.795))/0.026 = (3.26058×10^7 ± sqrt(1.06314×10^15))/0.026 = (3.26058×10^7 ± 3.26058×10^7)/0.026 = 1.25407×10^9 ± 1.25407×10^9:
x = 2.50814×10^9 or x = 0.0000259338

0.013 (x + 237.3) (x + 273.2) ⇒ 0.013 (237.3 + 0.0000259338) (273.2 + 0.0000259338) ≈ 842.795
3.26058×10^7 x ⇒ 3.26058×10^7 0.0000259338 ≈ 845.59:
So this solution is incorrect

0.013 (x + 237.3) (x + 273.2) ⇒ 0.013 (237.3 + 2.50814×10^9) (273.2 + 2.50814×10^9) ≈ 8.17797×10^16
3.26058×10^7 x ⇒ 3.26058×10^7 2.50814×10^9 ≈ 8.17797×10^16:
So this solution is incorrect

All the solutions are incorrect:
Answer: | NO SOLUTIONS!!

sum 1/2^(n-1), n=1

Sum of: 1/2^(1 - 1) =

Sum of: 1/2^0 =1

How many times does 4 go into 9?

if you have that many zeros, * 3,200, we would drop them and just multiply 1000* 3,200, I beleive.

3200*1000 = 3200. 

Find the greatest common divisor:
gcd(16, 20)
Find the divisors of each integer and select the largest element they have in common:
The divisors of 16 are:
1, 2, 4, 8, 16
The divisors of 20 are:
1, 2, 4, 5, 10, 20
The largest number common to both divisor lists is 4:
divisors of 16: 1, 2, 4, 8, 16
divisors of 20: 1, 2, 4, 5, 10, 20
Answer: |gcd(16, 20) = 4

A scuba diver standing on a boat is at an altitude of 1.3 meter above sea level. the scuba diver jumps into the water and decreases his altitude by 5.6 meter on one minute which equation can be used to determine the scuba divers altitude in the meter relative to sea level one minute after jumping into the water?

I get the emjority of the question, but it asks wich equation, in that case, their should be some opptions, I'm pretty sure!

You guys are hilarious!

365000000 days.

What is 2.83815852e16 kilometers in decibel number?

It is already in "Decimal form", but is written in "scientific notation". It can also be written as:

2.83815852 x 10^16.

Solve for m:
6/m^3 = 18

Take the reciprocal of both sides:
m^3/6 = 1/18

Multiply both sides by 6:
m^3 = 1/3

Taking cube roots gives 1/3^(1/3) times the third roots of unity:
Answer: |m = -(-1/3)^(1/3)          or m = 1/3^(1/3)          or m = (-1)^(2/3)/3^(1/3)

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

Formula:

\(\begin{array}{|rcll|} \hline \sin(10^{\circ}) = \cos(90^{\circ}-10^{\circ}) = \cos(80^{\circ}) \\ \sin(50^{\circ}) = \cos(90^{\circ}-50^{\circ}) = \cos(40^{\circ}) \\ \sin(70^{\circ}) = \cos(90^{\circ}-70^{\circ}) = \cos(20^{\circ}) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && csc(10^{\circ})+csc(50^{\circ})-csc(70^{\circ}) \\ &=& \frac{1}{\sin(10^{\circ})} + \frac{1}{\sin(50^{\circ})} - \frac{1}{\sin(70^{\circ})} \\ &=& \frac{1}{\cos(80^{\circ})} + \frac{1}{\cos(40^{\circ})} - \frac{1}{\cos(20^{\circ})} \\ &=& \frac{\cos(20^{\circ})\cos(40^{\circ})+\cos(20^{\circ})\cos(80^{\circ})-\cos(40^{\circ})\cos(80^{\circ})} {\cos(20^{\circ})\cos(40^{\circ})\cos(80^{\circ})} \\\\ && \cos(20^{\circ})\cos(40^{\circ})\cos(80^{\circ}) \\ &&= \frac{ 2\cdot\sin(20^{\circ}) \cos(20^{\circ})\cos(40^{\circ})\cos(80^{\circ})} {2\cdot\sin(20^{\circ}) } \\ &&= \frac{ \sin(40^{\circ})\cos(40^{\circ})\cos(80^{\circ})} {2\cdot\sin(20^{\circ}) } \\ &&= \frac{ \sin(80^{\circ})\cos(80^{\circ})} {4\cdot\sin(20^{\circ}) } \\ &&= \frac{ \sin(160^{\circ}) } {8\cdot\sin(20^{\circ}) } \\ &&= \frac{ \sin(20^{\circ}) } {8\cdot\sin(20^{\circ}) } \\ &&= \frac{ 1 } {8} \\\\ &=& 8\cdot [~ \cos(20^{\circ})\cos(40^{\circ})+\cos(20^{\circ})\cos(80^{\circ})-\cos(40^{\circ})\cos(80^{\circ}) ~] \\ &=& 8\cdot \{~ \cos(20^{\circ})[\cos(40^{\circ})+\cos(80^{\circ})]-\cos(40^{\circ})\cos(80^{\circ}) ~\} \\\\ && \cos(40^{\circ}) = \cos(60^{\circ}-20^{\circ}) = \cos(60^{\circ})\cos(20^{\circ})+\sin(60^{\circ})\sin(20^{\circ}) \\ && \cos(80^{\circ}) = \cos(60^{\circ}+20^{\circ}) = \cos(60^{\circ})\cos(20^{\circ})-\sin(60^{\circ})\sin(20^{\circ}) \\ && \cos(40^{\circ})+\cos(80^{\circ}) = 2 \cos(60^{\circ})\cos(20^{\circ}) \\\\ &=& 8\cdot [~ \cos(20^{\circ})2 \cos(60^{\circ})\cos(20^{\circ})-\cos(40^{\circ})\cos(80^{\circ}) ~] \quad | \quad \cos(60^{\circ}) = \frac12\\ &=& 8\cdot [~ \cos(20^{\circ})\cos(20^{\circ})-\cos(40^{\circ})\cos(80^{\circ}) ~] \\\\ && \cos(40^{\circ}) = \cos(80^{\circ}-40^{\circ}) = \cos(80^{\circ})\cos(40^{\circ})+\sin(80^{\circ})\sin(40^{\circ}) \\ && -\cos(60^{\circ})=\cos(120^{\circ}) = \cos(80^{\circ}+40^{\circ}) = \cos(80^{\circ})\cos(40^{\circ})-\sin(80^{\circ})\sin(40^{\circ}) \\ && \cos(40^{\circ})-\cos(60^{\circ})= 2\cos(80^{\circ})\cos(40^{\circ}) \\\\ &=& 8\cdot \{~ \cos(20^{\circ})\cos(20^{\circ})-\frac12 \cdot [\cos(40^{\circ})-\cos(60^{\circ})] ~\} \\ &=& 8\cdot \{~ \cos(20^{\circ})\cos(20^{\circ})-\frac12 \cdot \cos(40^{\circ})+\frac12 \cos(60^{\circ}) ~\} \quad | \quad \cos(60^{\circ}) = \frac12 \\ &=& 8\cdot [~ \cos(20^{\circ})\cos(20^{\circ})-\frac12 \cdot \cos(40^{\circ})+\frac14 ~] \\\\ && 1 = \cos(20^{\circ}-20^{\circ}) = \cos(20^{\circ})\cos(20^{\circ})+\sin(20^{\circ})\sin(20^{\circ}) \\ && \cos(40^{\circ}) = \cos(20^{\circ}+20^{\circ}) = \cos(20^{\circ})\cos(20^{\circ})-\sin(20^{\circ})\sin(20^{\circ}) \\ && 1+\cos(40^{\circ}) = 2 \cos(20^{\circ})\cos(20^{\circ}) \\\\ &=& 8\cdot \{~ \frac12[1+\cos(40^{\circ})] -\frac12 \cdot \cos(40^{\circ})+\frac14 ~ \} \\ &=& 8\cdot [~ \frac12 + \frac12 \cdot \cos(40^{\circ}) -\frac12 \cdot \cos(40^{\circ})+\frac14 ~] \\ &=& 8\cdot (~ \frac12 +\frac14 ~) \\ &=& 8\cdot (~ \frac34 ~) \\ &=& \mathbf{6} \\ \hline \end{array} \)

PROOF:
cosec 10 + cosec 50 - cosec 70
sec 80 + sec 40 - sec 20
(1/cos 80) + (1/cos 40) - (1/cos 20)
(cos 40 + cos 80) / (cos 80 cos 40) - (1/cos 20)
Apply cos A + cos B = 2 cos ((A + B)/2) cos ((A - B)/2)
(2 cos 60 cos 20) / (cos 40 cos 80) - (1/cos 20)
(cos 20) / (cos 40 cos 80) - (1/cos 20)
Take LCM and add both
(cos²20 - cos 40 cos 80) / (cos 20 cos 40 cos 80)

I am gonna solve the denominator seperately,
cos 20 cos 40 cos 80
Multiply and divide by 2 sin 20
(2 sin 20 cos 20 cos 40 cos 80) / (2 sin 20)
2 sin A cos A = sin 2A
(sin 40 cos 40 cos 80) / (2 sin 20)
Multiply and divide by 2
(2 sin 40 cos 40 cos 80) / (4 sin 20)
(sin 80 cos 80) / (4 sin 20)
Again 2,
sin 160 / (8 sin 20)
sin (180 - 20) = sin 20
sin 160 = sin 20
= 1/8
So, denominator is equal to 1/8

Our expression becomes,
8 (cos²20 - cos 40 cos 80)
Take one 2 inside and use 2 cos A cos B = cos ((A + B)/2) + cos ((A - B)/2)
4 (2cos²20 - (cos 120 + cos 40))
2cos²A = 1 + cos 2A
4 (1 + cos 40 - cos 120 - cos 40)
4 (1 - cos 120)
4 (1 - (-1/2))
4 (3/2) = 6

Right, so:

27x³ - 8y³ = (3x)³ - (2y)³

\(\frac{\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)}{\left(3x-2y\right)}\)

= 9x² + 6xy + 4y²

Out of curiosity however, is it possible to use algebraic long division on something like this (with more than 1 variable?)

Thanks.

(27x3 - 8y3)/(3x-2y)

You do not ned to use algebraioc long division.  Use this :))

\(\boxed{a^3-b^3=(a-b)(a^2+ab+b^2)}\)

There can be sixteen different combinations

15% = 0.15 ??

(A x 0.15) + A = 27.850 

1.15A = 27.850

A = 27.850/1.15

A = 24.2173913043478261

compute 

1 - 2 + 3 - 4 + \(\dots \)+ 2005 - 2006 + 2007 = ?

Geometric series:

\(\begin{array}{|rcll|} \hline 1+x+x^2+x^3+x^4+x^5+\dots + x^n &=& \dfrac{1-x^{n+1}}{1-x} \qquad -1 < x < 1 \\ \hline \end{array} \)

Derivative:

\(\begin{array}{|rcll|} \hline 1+2x+3x^2+4x^3+5x^4+6x^5+\dots + nx^{n-1} &=& \dfrac{1-x^n[~n(1-x)+1~]}{(1-x)^2} \\ \hline \end{array} \)

x \(\rightarrow \) -x

\(\begin{array}{|rcll|} \hline 1-2x+3x^2-4x^3+5x^4-6x^5+-\dots + n(-x)^{n-1} &=& \dfrac{1-(-x)^n[~n(1+x)+1~]}{(1+x)^2} \\ \hline \end{array} \)

x = 1:

\(\begin{array}{|rcll|} \hline 1-2+3-4+5-6+-\dots + n(-1)^{n-1} &=& \dfrac{1-(-1)^n[~2n+1~]}{4} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \sum \limits_{n=1}^{2007}n (-1)^{n - 1} = 1 - 2 + 3 - 4 + \dots + 2005 - 2006 + 2007 \\\\ &=& \dfrac{1-(-1)^{2007}[~2\cdot2007+1~]}{4} \\\\ &=& \dfrac{1+4015}{4} \\\\ &=& \dfrac{4016}{4} \\\\ &\mathbf{=}& \mathbf{1004} \\ \hline \end{array} \)

e^pi*i

\(e^{\pi*i}=cos\pi+isin\pi=-1+i*0=-1\)

now using the web2.0 calc    :)

e^(pi*i) = -1