All Answers 358406 Answers

98 = 7^2 x 2 = 49 x 2

And 49 is the perfect square factor of 98

[-32 ± √72] / 4 =

[-32 ± √(36*2)] / 4 =

[-32 ± 6√2] / 4  =

[-16 ± 3√2] / 2

Here's A

2,  4,  8,  16,  32, 64, 128,   256,    512

Here's B

20,40,60,80,100,120,140,160,180,200,220,240,260,280, 300, 320

The least positive difference appears to be 4...this occurs three times

20 - 16,  64-60    and    260-256

It means "not equal to"

Thus ...    13 ≠ 14  ...!!!

So we have something like this...

1  2  4  8  16  32   64   128  

3

9

27

81

243

729

2187

Let's see if we can discover a pattern.....!!!

For the first row the sum is just 1(1-2^8)/(1-2)  = 255

For the second row, the sum is just 3(1-2^8)/(1-2) = 765

That leads us to believe that the sum of all numbers on the chessboard is given by the sum of the numbers in the first column times (1-2^8)/(1-2)

And the sum of the numbers in the first column is just 1(1-3^8)/(1-3) = 3280

So the total sum of numbers on the chess board is given by 3280(1-2^8)/(1-2) = 836,400...!!!

Another way to look at this is to notice that every successive row sum is triple the previous one

So we would have  255(1-3^8)/(1-3) = 836,400...just as we expected..!!!

thanks alot

Here's another possibility...

Notice that the 2nd and 3rd terms sum to 1

And the 4th and 5th terms sum to 2/3

And the 6th and 7th terms sum to 4/9

So we have....adding in the first term.....

1 + 1/[1-(2/3)] = 1 + 1/(1/3)  = 1 + 3  = 4

Here's c

We have to use the Law of Cosines, first....let's find angle P....so we have

25^2  = 18^2 + 16^2 - 2(18)(16)cosP   simplify

625 = 324 + 256 - 576cosP      subtract 324 and 256 from both sides and divide both sides by - 576

(625 - 324 - 256) / (-576) = cosP.....and we can use the inverse cosine to find P

cos^-1 (625 - 324 - 256) / (-576) = P = about 94.48 degrees

And now, we can use the Law of Sines to find angle R

So we have

sinR/16 = sinP/25

sinR/16 = sin(94.48)/25   multiply both sides by 16

sinR = 16*sin (94.48)/25  = .638

And using the sine inverse we have

sin^-1(.638) = about 39.64 degrees

And the remaining angle, Q = 180- 94.48 - 39.64 = about 45.88 degrees

This makes sense....the greatest angle is opposite the greatest side, the next greatest angle is opposite the next greatest side, and the smallest angle is opposite the smallest side...

Here's b

Since the sum of any three angles of a triangle = 180....then angle X must just be given by

X + 70 + 40 = 180    simplify

X + 110 = 180          subtract 110 from both sides

X = 70

And using the Law of Sines we can find side "y"...so we have

12 / sin40  = y/ sin 70     multiply both sides by sin70 and we have

12*sin70 / sin40 = y = about 17.54

Also the remaining side is the same length because angle X and Y are equal...and in a triangle....sides opposite equal angles are themselves equal..!!!

Thus....this is an isosceles triangle....(two equal sides)

Notice that the first and last terms = 2008

And notice that the second term and the next-to-the-last term  = -2 - 2206 = -2008

And there are 1002 pair-wise additions like this that will sum to "0"

That leaves three "middle" terms, 1003 - 1004 + 1005 = 1004....and that's the sum

would be grateful if someone could explain how to do these, have test on this topic tomorrow and i've been abscent for 2 weeks

Let's see if we can discover a pattern....

1/3 + 1/12 + 1/30 + 1/60 + 1/105.....

1/(3*1) + 1/(6*2) + 1/(10*3) + 1/(15*4) + 1/(21*5)......

The form is

∑1/((n)(n+1)(n-1)/2) = ∑2/((n)(n+1)(n-1)) for n=2 to n=infinity

And WolframAlpha  gives this sum as  1/2

Do this as 2 seperate APs and then add the sums together. 

3/2 - 5k = -2(4-k)

3-10k     =  -4(4-k)

3-10k      =  -16+4k

19            =   14k

            k  =  19/14

$$\\8x^2-x^3=70\\\\
-8x^2+x^3=-70\\\\
x^3-8x^2+70=0\\\\$$

GP

a=1/3     r= 1/3     n=6

$$\\\boxed{S_n=\frac{a(1-r^n)}{1-r}}\\\\
S_6=\frac{\frac{1}{3}(1-(\frac{1}{3})^6)}{1-\frac{1}{3}}}\\\\
S_6=\frac{\frac{1}{3}(1-\frac{1}{3^6})}{\frac{2}{3}}}\\\\
S_6=\frac{1}{\not{3}}(1-\frac{1}{3^6})\times \frac{\not{3}}{2}\\\\
S_6=\frac{1}{2}(1-\frac{1}{3^6})\\\\$$

$${\mathtt{0.5}}{\mathtt{\,\times\,}}\left({\mathtt{1}}{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{{\mathtt{3}}}^{{\mathtt{6}}}}}\right)\right) = {\frac{{\mathtt{364}}}{{\mathtt{729}}}} = {\mathtt{0.499\: \!314\: \!128\: \!943\: \!758\: \!6}}$$

Again, we are both correct but heureka's method is preferable.        Mine is pretty silly really   

Thanks Heureka      

There are 2 GPs here

$$\\\[ 1+\frac{2}{3}+\frac{4}{9}+\frac{8}{81}+\dotsb \]\\\\
S_{\infty}=\frac{a}{1-r}=\frac{1}{1-\frac{2}{3}}=\frac{1}{\frac{1}{3}}=3\\\\$$

-----------------------------------------

$$\\\[ \frac{1}{3}+\frac{2}{9}+\frac{4}{27}+\frac{8}{81}+\dotsb \]\\\\
S_{\infty}=\frac{a}{1-r}=\frac{\frac{1}{3}}{1-\frac{2}{3}}
=\frac{\frac{1}{3}}{\frac{1}{3}}=1\\\\\\
Total=3+1=4$$

$$\small{\text{
sum
$
s=
1+1*( \frac{2}{3} ) ^0+
+1*(\frac{2}{3})^1+
+1*(\frac{2}{3})^2+
+1*(\frac{2}{3})^3+
+1*(\frac{2}{3})^4+
\dots
$
}}\\
a= 1 \\
r = \frac{2}{3} \\
s = 1 + \frac{a}{1-r} = 1 + \frac{1}{1-\frac{2}{3}} = 1 + \frac{1} {\frac{1}{3}} = 1 + 3 = 4 \\
\boxed{s= 4}$$

really melody it hasn't a meaning. so we can't solve this problem.

12* x =-12

x   =-12/12

x   = -1//

that is the answer.

Thanks Chris,  

Yes, I will admit, I like Heureka's method better too.   

Also well done, heureka...!!!

Your method is a little more intuitive to me, than Melody's.....

But.......either one gets the job done!!!