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Simplify
\(\dfrac{1}{1 + \sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} + \dfrac{1}{\sqrt{3} + \sqrt{4}} + \dots + \dfrac{1}{\sqrt{99} + \sqrt{100}}\).

\(\small{ \begin{array}{|rcll|} \hline &&\mathbf{\dfrac{1}{1 + \sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} + \dfrac{1}{\sqrt{3} + \sqrt{4}} + \dots + \dfrac{1}{\sqrt{99} + \sqrt{100}}} \\\\ &=& \dfrac{1 - \sqrt{2}} {(1 + \sqrt{2})(1 - \sqrt{2})} + \dfrac{\sqrt{2} - \sqrt{3}} {(\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3})} + \dfrac{\sqrt{3} - \sqrt{4}} {(\sqrt{3} + \sqrt{4})(\sqrt{3} - \sqrt{4})} + \dots + \dfrac{\sqrt{99} - \sqrt{100}} {(\sqrt{99} + \sqrt{100})(\sqrt{99} - \sqrt{100})} \\\\ &=& \dfrac{1 - \sqrt{2}} {1-2} + \dfrac{\sqrt{2} - \sqrt{3}} {2-3} + \dfrac{\sqrt{3} - \sqrt{4}} {3-4} + \dots + \dfrac{\sqrt{99} - \sqrt{100}} {99-100} \\\\ &=& \dfrac{1 - \sqrt{2}} {-1} + \dfrac{\sqrt{2} - \sqrt{3}} {-1} + \dfrac{\sqrt{3} - \sqrt{4}} {-1} + \dots + \dfrac{\sqrt{99} - \sqrt{100}} {-1} \\\\ &=& -1 +\underbrace{ \sqrt{2} -\sqrt{2} + \sqrt{3} -\sqrt{3} + \sqrt{4} + \dots -\sqrt{99}}_{=0} + \sqrt{100} \\\\ &=& -1 + \sqrt{100} \\\\ &=& -1 + 10 \\\\ &=& \mathbf{ 9 } \\ \hline \end{array} }\)

\(\dfrac{1}{1 + \sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} + \dfrac{1}{\sqrt{3} + \sqrt{4}} + \dots + \dfrac{1}{\sqrt{99} + \sqrt{100}} = \mathbf{ 9}\)

The area of triangle ABD is 144 cm^2.

The area of the circle is 125*pi.

The inequality is x^2 - 4x + 3 > 0, so b + c = (-4) + 3 = -1.

\(87= 21^x\\ ln87= ln21^x\\ ln87=x ln21\\ x=\frac{ln87}{ln21}\)

x= ln(87)/ln(21) = approx 1.4668665480402957

The area of triangle PQR is 1/2*6^2 = 18.

Angle JGH = 180 - 2 * 62 = 56 degrees

In the diagram below, ABCD is a parallelogram with AB=42 cm. Arc AD is a semi-circle. The area of the parallelogram is 504cm2, and angle ADC = 60 degrees. Find the area of the shaded region in cm2.

AB = 42cm       

Area = 504 cm²

Angle ADC = 60°

504 / 42 = 12

Semi-circle radius  r =  [12 / sin(60°)] /2    r = 6.928

Shaded region area    A_r = ( r² pi ) /2 = 75.398 cm²    

Let the shorter leg =S
The longer leg =5S + 19
Hypotenuse      =5S + 20
[5S + 20]^2 =[5S + 19]^2 + S^2, solve for S
S = 13 inches - the shorter leg
13 * 5 + 19 =84 inches - the longer side
13 *5 + 20  =85 inches - the hypotenuse
85^2 =84^2 + 13^2 - which balances. 

Moana's age now =M
Pedro's age now  =2M

2M - 4 =4(M - 4), solve for M
M = 6 years - Moana's age now
6 x 2 =12 years - Pedro's age now

Probability of getting 2 sixes:

D=8; C=2; ((D nCr C * 5^(D-C)) / 6^D = 26.05 %

Could you also please answer this question?

The sum of the first $n$ terms of a certain sequence is $n(n + 1)(n + 2).$ Find the tenth term of the sequence.

The computer simply sums up all these fractions and get 3/2:

0.75
0.25
0.125
0.075
0.05
0.0357142857 1
0.0267857142 9
0.0208333333 3
0.0166666666 7
0.0136363636 4
.

.

............etc.

Let the radius of M  =   m

We can from a right triangle

Two  legs = (m - 5/ [ 1 + sqrt (2) ]

And the  hypotenuse =  m + 5 / [ 1 + sqrt (2) ]

So.....by the pythagorean Theorem we have that

(m - 5/ [ 1 + sqrt (2) ] )^2   +  ( m - 5/ [1 + sqrt (2) ] )^2   = ( m  + 5 / [1 + sqrt (2) ] )^2

Solving this  for the larger value of m  we get that   m   =  5 ( 1 + sqrt (2) ) ≈  12.07 m

Pounds of high-quality coffee =H

Pounds of cheaper coffee        =190 - H

5.50H + 3.25[190 - H] =190*4.55, solve for H

H = ~110 pounds of high-quality cofffee

190 - 110 = 80 pounds of cheaper coffee.

Can you please explain how you got to that answer, and it is correct btw thanks!

How many four-letter words can be made from the letters "TOMORROW"?

I assume nonsense works are ok?  I expect you just want 4 letter strings.

TMW OOO RR

I think it is just    8P4/( 3!2!) = 1680/12 = 140

sum_(n=1)^∞ 6/((2 n + 1)^2 - 1) = 3/2

The side of the equilateral triangle, S,   will   =   the radius of the  circle

Area of equilateral  triangle  = (sqrt(3) /4) S^2

10  = (sqrt (3) / 4 ) * S^2

10 * 4  /sqrt (3)  =  S^2

40 / sqrt (3)  = S^2       take the square  root  of both sides

S =  sqrt (40) / (3)^(1/4)  = S =  circle radius

arc length =  2pi  * R *  [ central angle measure /  360)  =      2pi * [ sqrt (40 / 3^(1/4) ]   (111/ 360 ) ≈    9.31 cm

1 quarter =0.25

3 dimes   =0.30

7 nickles  =0.35

10 pennies=0.10

21 coins      = $1

Note that OL  and  ON  form a 90° angle because they are perpendicular

This must mean that the sum of angles LOM and MON  = 90°   ........so.....

3x -  15  + 5x  -  23   =   90      simplify

8x - 38  = 90     add 38  to  both sides

8x =  128         divide  both sides  by 8

x = 16

So....MON =   5(16)  - 23 =   80  - 23   =  57°

1 - The ratio of the minute hand to that traveled by the hour hand is 12:1. Since the distance traveled around the dial by the minute hand is 12/13 of a revolution and the distance traveled by the hour hand is 1/13 of a revolution, then: 12/13 x 60 =55 5/13 minutes before noon.

2 - Since the minute hand travels 12 times faster than the hour hand, then the distance between them was 11/12 of the distance of the minute hand from 12:00 noon. Hence the time the business man left for lunch is 11/12 of the distance, which is 1/13 of a revolution. Therefore, the distance is:12/11 x 1/13 x 60=5 5/143 minutes past noon. It then follows that the time when he came back is:5 5/143 + 55 5/13 =60 60/143. That is: 60/143 of a minute after 1:00 pm.