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I wil assume   'clear the fence'   means 'clear the net' AND she is hitting the ball PERPINDICULAR to the net from 39 feet (468 inches) and not diagonal across the court.....    I will also assume she hits the ball from a height equal to her height PLUS the racquet length of 27 "   =  94 inches (sorry for all of the assumptions. but this info is not included in the Question)   I will also have to assume she hits the ball horizontally with no VERTICAL velocity....Also have to assume the net is 36 inches HIGH (not long)

Ball position in the horizontal distance is equal to

x = xo + vot + 1/2 a/t^2     (1)

Let's do the 'y' component first

36 inches = 94 inches  + (0)t + 1/2 a t^2       t = 1.898 secs    (vo in the VERTICAL direction is zero since she hits the ball horizontally)

Now for the 'x'  (horizontal )  calcs:

39 ft (12 in/ft) =  0 + vot + 1/2 a t^2        For this   a = 0 (negleting air friction)

39(12) = vo t       and t = 1.898 sec        so vo = 246.57 in/sec = 20.54 f/s    

Thanks Chris, I believe your answer is correct but I'd like to talk about it a different way.

Let the people form a line and the envelopes are on the table.

The first person has a 1 in 5 chance of selecting the right letter.  (He selects the correct one)

Now there are 4 on the table so the second person has a 1 in 4 chance of getting the right one (He selects the correct one)

Now there are 3on the table so the second person has a 1 in 3 chance of getting the right one (He selects the correct one)

and so on

so 

P(everyone getting the correct letter) = \(\frac{1}{5}\times \frac{1}{4}\times \frac{1}{3}\times \frac{1}{2}\times \frac{1}{1}=\frac{1}{5!}=\frac{1}{120}\)

I think this might be correct......but maybe someone else has a better solution

Call the set of  people  A  B  C  D  E

And call the set of  the pieces of mail A B C D E in that order

Note that there are  5!  different "words" that can be formed by the set  of letters in the first set

But....only one of these,  ABCDE,  matches the second set

So....the probability that they all get the correct pieces of mail is   1 / 120 

FV (future value) is an importtant value that needs to be preserved. The top part of  the formula may be zero but not always.

the software it is going in has to solve for all values. i have 2/3 but cant figure out i and n

Similarly:

#teams left = 128/(2^x)      where x = rounds

16 = 128/(2^x)

128/16 = 2 ^x

log (128/16) / log 2  = x       x = 3 rounds

the number of teams after round  0   =   128

the number of teams after round  1   =   (1/2) * 128

the number of teams after round  2   =   (1/2) * (1/2) * 128

the number of teams after round  x   =    (1/2)^x * 128

(1/2)^x * 128   =   16

                                     Divide both sides of the equation by  128 .

(1/2)^x   =   1/8

                                     We can express  1/8  as  (1/2)³

(1/2)^x   =   (1/2)³

                                     So we can see that...

x  =  3

Since it's even, we can divide by 2

576  / 2  =   288 

288 / 2   =  144

144 / 2  =  72

72 / 2  =  36

36 / 2 = 18

18 / 2  = 9

9  =  3 * 3

Take  the terms in red and we have

2⁶  * 3²

Ah, you are correct, hectictar....I missed the word "congruent '

The point  (1,1) will be on any graph of this type....and  the range ⇒   inf   as  x ⇒ inf

So...the range of these functions will be  [ 1, inf )

Wait a minute!!!

A. Trapezoid EFGH is not congruent to trapezoid E′F′G′H′ because there is no sequence of rigid motions that maps trapezoid EFGH to trapezoid E′F′G′H′ .

thank you

Here's how I like to think of these...

original perimeter   =   2(L + W)

To scale the photo, mulitply the length by  1.75  and the width by  1.75 , so that

the new length  =  1.75L   and   the new width  =  1.75W

new perimeter   =   2(1.75L + 1.75W)

new perimeter   =   1.75 * 2(L + W)

new perimeter   =   1.75 * original perimeter

new perimiter    is   1.75 times the original perimeter

original area   =   L * W

new area   =   1.75L * 1.75W

new area   =   1.75 * 1.75 * L * W

new area   =   1.75² * L * W

new area   =   1.75² * original area

new area   =   3.0625 * original area

new area   is   3.0625 times the original area

8 m  increased by  25%  of  8 m

=  8 m  +  25%  of  8 m

=  8 m  +  0.25 * 8 m

=  1.25 *  8 m

=  10 m

a)  We can use the law of cosines to find the angles.

c²   =   a² + b² - 2ab cos C    , where  C  is the angle opposite side  c .

Let  a = 19 ,  b = 20 ,  c = 21  and  A ,  B , and  C  be the angles opposite their respective sides.

21²   =   19² + 20² - 2(19)(20)cos C

441   =   361 + 400 - 760 cos C

441   =   761 - 760 cos C

                                                     Subtract  761  from both sides of the equation.

-320   =   -760 cos C

                                                     Divide both sides by  -760 .

8/19   =   cos C

                                                     Take the inverse cosine of both sides.

C  =   acos( 8/19 )   ≈   65.099°

Now let's use the law of cosines again to find the angle opposite the  20 cm  side.

20²   =   19² + 21² - 2(19)(21)cos B

400   =   361 + 441 - 798 cos B

400   =   802 - 798 cos B

-402   =   -798 cos B

67/133   =   cos B

B   =   acos( 67/133 )   ≈   59.751°

Since there are  180°  in every triangle,  A + B + C  =   180°   and   A  =  180° - B - C

A   ≈   180° - 59.751° - 65.099°   ≈   55.15°

b)

Let the base be the 19 cm side, and the height be  20 sin 65.099°

area   ≈   (1/2)(19)(20 sin 65.099°)   ≈   172.337  sq cm

C. Trapezoid EFGH is congruent to trapezoid E′F′G′H′ because you can map trapezoid EFGH to trapezoid E′F′G′H′ by dilating it by a factor of 3/4 and then translating it 2 units left, which is a sequence of rigid motions.

Acute scalene triangle.

Sides: a = 19   b = 20   c = 21

Area: T = 172.3376879396
Perimeter: p = 60
Semiperimeter: s = 30

Angle ∠ A = α = 55.1550095421° = 55°9' = 0.96325507479 rad
Angle ∠ B = β = 59.75109669494° = 59°45'3″ = 1.04328511045 rad
Angle ∠ C = γ = 65.09989376296° = 65°5'56″ = 1.13661908012 rad

Height: ha = 18.1410724147
Height: hb = 17.23436879396
Height: hc = 16.4133036133

Median: ma = 18.17327818454
Median: mb = 17.34993515729
Median: mc = 16.43992822228

Inradius: r = 5.74545626465
Circumradius: R = 11.57661641211

Vertex coordinates: A[21; 0] B[0; 0] C[9.57114285714; 16.4133036133]
Centroid: CG[10.19904761905; 5.47110120443]
Coordinates of the circumcenter: U[10.5; 4.87441743668]

https://www.triangle-calculator.com/?what=sss&a=19&b=20&c=21&submit=Solve

"a" is an alternate interior angle  to the angle measuring 36°....so  that is its measure

By the Exterior Angle Theorem, "c"  =  139 - 77  =  62°

We can stop here....there is only one answer where a = 36  and c  = 62

(2/3)x  + (3/5)y  =  12

-3x   + (5/2)y  =  6

Mutiply  the first equation through by the common denominator of 3 and 5 = 15

Multiply the second equation through by 2

So we have

10x  +  9y   =  180    multipy through by 6  =   60x + 54y  = 1080    (1)

-6x  +  5y      =  12  multiply through   by 10  =   -60x  + 50y  =   120    (2)

Add (1)  and (2)

104y  =  1200     divide by  104

y =   1200/104 =   600/52  =  150 / 13

Obviously, only one answer has a "y" answer that  =  150/13.....so that must be the correct one

your welcome darkfox

Notice that   angle ABE and FDE are alternate interior angles and thus, they are equal

And angle CDE  is supplemental to FDE

So....

CDE  + FDE  =  180

CDE   +  30  =   180     subtract 30 from both sides

CDE  =   150   =  "y"

Sorry if it's such a big question

Thank you, Foxy. Have a good day!

They are alternate interior angles.