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Calculate:

\(\large{\frac{1}{11*15}+\frac{1}{19*15}+\frac{1}{19*23}+\frac{1}{23*27}+\cdots+\frac{1}{1103*1107}+\frac{1}{1111*1107}}\)

\(\begin{array}{|rcll|} \hline && \frac{1}{11*15}+\frac{1}{15*19}+\frac{1}{19*23}+\frac{1}{23*27}+\cdots+\frac{1}{1103*1107}+\frac{1}{1107*1111} \\\\ &=& \frac14\left( \frac{1}{11}-\frac{1}{15}\right) +\frac14\left( \frac{1}{15}-\frac{1}{19}\right) +\frac14\left( \frac{1}{19}-\frac{1}{23}\right) +\frac14\left( \frac{1}{23}-\frac{1}{27}\right) +\cdots\\ && +\frac14\left( \frac{1}{1103}-\frac{1}{1107}\right) +\frac14\left( \frac{1}{1107} -\frac{1}{1111} \right)\\\\ &=& \frac{1}{4*11} \\ &&-\frac14\left(\frac{1}{15}\right)+ \frac14\left(\frac{1}{15}\right)\\ &&-\frac14\left(\frac{1}{19}\right)+\frac14\left(\frac{1}{19}\right) \\ &&-\frac14\left(\frac{1}{23}\right)+\frac14\left(\frac{1}{23}\right) \\ &&-\frac14\left( \frac{1}{27}\right)+\frac14\left( \frac{1}{27}\right)+\cdots \\ &&-\frac14\left( \frac{1}{1103}\right)+\frac14\left( \frac{1}{1103}\right)\\ &&-\frac14\left( \frac{1}{1107}\right)+\frac14\left( \frac{1}{1107}\right) \\ &&-\frac14\left( \frac{1}{1111} \right)\\\\ &=& \frac{1}{44}-\frac14\left( \frac{1}{1111} \right) \\\\ &=& \frac{1}{44}- \frac{1}{4444} \\\\ &\mathbf{=}& \mathbf{\frac{275}{12221}} \\\\ &\mathbf{=}& \mathbf{\frac{25}{1111}} \\\\ &\mathbf{=}& \mathbf{0.\overline{0225}} \\ \hline \end{array}\)

Calculate:

\(\large{\frac{1}{2*4}+\frac{1}{6*4}+\frac{1}{6*8}+\frac{1}{10*8}+\cdots+\frac{1}{96*98}+\frac{1}{100*98}}\)

\(\begin{array}{|rcll|} \hline && \frac{1}{2*4}+\frac{1}{4*6}+\frac{1}{6*8}+\frac{1}{8*10}+\frac{1}{10*12}+\frac{1}{12*14}+\cdots+\frac{1}{94*96}+\frac{1}{96*98}+\frac{1}{98*100} \\\\ &=& \frac12\left( \frac{1}{2}-\frac{1}{4}\right) +\frac12\left( \frac{1}{4}-\frac{1}{6}\right) +\frac12\left( \frac{1}{6}-\frac{1}{8}\right) +\frac12\left( \frac{1}{8}-\frac{1}{10}\right) \\ &&+\frac12\left( \frac{1}{10}-\frac{1}{12}\right) +\frac12\left( \frac{1}{12}-\frac{1}{14}\right)+\cdots\\ && +\frac12\left( \frac{1}{94}-\frac{1}{96}\right) +\frac12\left( \frac{1}{96}-\frac{1}{98}\right) +\frac12\left( \frac{1}{98} -\frac{1}{100} \right)\\\\ &=& \frac14 \\ &&-\frac12\left(\frac{1}{4}\right)+ \frac12\left(\frac{1}{4}\right)\\ &&-\frac12\left(\frac{1}{6}\right)+\frac12\left(\frac{1}{6}\right) \\ &&-\frac12\left(\frac{1}{8}\right)+\frac12\left(\frac{1}{8}\right) \\ &&-\frac12\left( \frac{1}{10}\right)+\frac12\left( \frac{1}{10}\right) \\ &&-\frac12\left( \frac{1}{12}\right)+\frac12\left( \frac{1}{12}\right) \\ &&-\frac12\left( \frac{1}{14}\right)+\frac12\left( \frac{1}{14}\right)+\cdots \\ &&-\frac12\left( \frac{1}{94}\right)+\frac12\left( \frac{1}{94}\right)\\ &&-\frac12\left( \frac{1}{96}\right)+\frac12\left( \frac{1}{96}\right) \\ &&-\frac12\left( \frac{1}{98}\right)+\frac12\left( \frac{1}{98}\right) \\ &&-\frac12\left( \frac{1}{100} \right)\\\\ &=& \frac14-\frac12\left( \frac{1}{100} \right) \\\\ &=& \frac14- \frac{1}{200} \\\\ &\mathbf{=}& \mathbf{\frac{49}{200}} \\\\ &\mathbf{=}& \mathbf{0.245} \\ \hline \end{array}\)

Hi Alan thanks for your follow up......If I may bother you for a bit more clarification

>>The formulas i am following are as below:

Therefore, I have understood solutions obtained as far as angles go...

but

the logic of how you derived  extensions (as highlighted below in your answer) escapes me.

Thanks for your patience!

\(\text{divisible por (x-2) significa p(2) = 0}\\ p(2) = 2(8) - 9(4) + 13(2) + k = \\ 16-36+26+k =\\ 6+k = 0\\ k = -6\)

C other post

This is false.  Consider c < 0

\(\text{The first statement is false}\\ |z| \in \mathbb{R},~\forall z \in \mathbb{C}\\ i |z| \not \in \mathbb{R}\)

\(|z \cdot w | =\\ |(z_r w_r -z_i w_i) + i(z_r w_i + z_i w_r)| = \\ \sqrt{(z_r w_r -z_i w_i)^2 + (z_r w_i + z_i w_r)^2} = \\ \sqrt{-2 w_i z_i w_r z_r+w_i^2 z_i^2+w_r^2 z_r^2 + z_i^2 w_r^2+2 w_i z_i w_r z_r+w_i^2 z_r^2}=\)

\(\sqrt{w_r^2 z_r^2 + z_r^2 w_i^2 + z_i^2 w_r^2 + w_r^2 w_i^2}=\\ \sqrt{(z_r^2+z_i^2)(w_r^2+w_i)^2} = \\ \sqrt{z_r^2 + z_i^2}\sqrt{w_r^2+w_i^2} =\\ |z||w|\)

oh it says integer lengths... DOH

\(3 < x < 3 \sqrt{5} \approx 6.7\\ \text{That gives you 4, 5, and 6}\)

4 gives you an angle of about 36.36 degrees

5 gives you an angle of about 56.25 degrees

6 gives you an angle of about 75.53 degrees

all 3 are acute triangles

Sorry....that should have been 210 gm....I have corrected it......THANX !

X = price the STORE pays for instr.

130% X   = $ 104

X = 104/(130%) = 104/1.3 = $ 80

1 x 6 = 6

3 x 2 = 6       You have to paint one number from EACH pair

So you have to paint     1  3       1   2       6  3        6 2

4 ways

Thank you so much! I'm pretty good at understanding chemistry, but I can't do math for the life of me. 

Edit: In your answer you listed the total mass as 220 grams instead of 210. Is that correct?

d= kt^2

80 = k (4)^2

k=5

d=(5)(t^2)   in the first TWELVE SECONDS

d = 5(12^2) = 720 m   

     MINUS the distance it fell in the FIRST 4 seconds 

          720 - 80 = 640 m

1. Is it possible to inform me on how these solutions are so obvious? 

2. My bad, I'm sorry, I fixed it though 

The center is (0,0)

The since (8,2) is on the graph....the equation of the asymptote will have a  positive slope...so....

y = (b / a) x        and we know that  (x, y)  = (8, 2)....so....

2 = (b / a) (8)

2/8 = (b / a)

1/4 =  b / a        which implies that

4 /1  =  a / b   =   4

1)

x^2 + y^2=x^3
5^2 + 10^2=5^3
10^2  + 30^2 =10^3

Well....that 'ain't' much ice......it will warm up and then melt at constant temp and cool the entire mass (200 + 10gm) of a bit

You'll need     Specific heat of ice (0.5 cal /gm-C   or    2093 j/kg-c )

10 gm = .01 kg ice

.01 kg (2093 j/kg-c)(5 C)  = 104.65 j   to warm the ice to 0 DEGREES

Now you'll need the latent heat of fusion of ice to melt it    .336 MJ/kg

.01kg (.336 Mj/kg) = 3360 j

The heat to warm the ice and melt it has to come from the water

 the final mass of water ( 200+10)     will cool off an amount in joules equal to the two amounts we just calc'd

So, you'll need      Specific heat of water  4.186 j/gm c

210gm (4.186 j/gm-c) (45-x c ) = 104.65 + 3360 j

45-x = 3.9413 (the water loses 3.9413 degrees)

41.059  degrees C is equalibrium

       Corected....changed 220 to 210 in last calcs.....

Close...

We can obtain the graph of \(y=g(x)\) by taking the graph of \(y=f(x)\) and stretching it horizontally by a factor of 2, then shifting it down by 4 units. Therefore, \(g(x) = \boxed{f \left( \frac{x}{2} \right) - 4}.\)

Isn't this correct?

\(\lfloor x \rfloor = -9 \Rightarrow x \in [-9,-8)\\ 5x \in [-45,-40)\\ \text{so }\lfloor 5x \rfloor \in \{-45, -44, -43, -42, -41\}\\ \text{5 possible values}\)

\(\text{The list of officers can be considered a 3 digit base 25 number}\\ N = (25)^3 = 15625\)

Ah ok i get it now, i wrote MR= 20 thanks!

it's B

OK, YEEEEEET....I think the radius should be sqrt 225 = 15

Remember that if we draw a perpendicular to the chord from A....this perpendicular will bisect the chord.......call the point where this perpendicular intersects the chord, M

And we will have a right triangle   AMR

MR = 10

AR = 15 = the hypotenuse of AMR

The other leg, AM....is the distance we seek

So

AM =  sqrt [ (15 )^2 - 10^2 ]  =  sqrt [ 225 - 100] = sqrt (125) = 5sqrt(5)

um b?