CPhill

1.  We can generalize the true side lengths of GHI  and the hexagon from the following pic, ACG

GHI  will form an equilateral triangle....call the side, S

Its  area  =   (1/2) S^2  sin ( 60) =   (1/2)S^2√3/2   =   √3/4S^2

So....we can find the length of one side of GHI, thusly

225  = √3/4 S^2

225 * 4 / √3   = S^2

900 / √3   =  S^2

( 30 /⁴√3  )  = S

Look at the pic.....triangle CJB  comprises 1/6 of the hexagon's area....and it is an equilateral triangle with a side of 2

And note that in the pic....its side will be (2/3)  that  of the true side length of triangle GHI   = 20/⁴√3

So....the area of the hexagon is just

(6) (1/2) (20/⁴√3)^2 sin 60  =

3 (400/√3)√3/2   =

(3/2)*400  =  600 units^2

Second one  ;

M + T + H  + A  + S  = 540

M + T + H + A  + (180-A)  = 540

Since M = T = H  we have that

3H + A  + 180 - A  = 540

3H + 180  = 540

3H  = 360

H  = 120°

N  + .20N   = 90  simplify

1.20N   = 90       divide both sides by 1.20

N  = 75  = the number you started with

Verify

75  +  20%  of 75  =

75  +  (20/100)*75  =

75  +  (1/5)*75  =

75  +  15  =

90  ....

foci  (0, - 5)    and (0, 5)

vertices (0, 4)  and (0, -4)

Since the vertices and focus are on the y axis  and the center is (0,0).....this hperbola will have the form

y^2          x^2

___  -     ____    =  1

a^2          b^2

a  = 4

c  =5

and b  = √[5^2  - 4^2 ]  =  3

So....we have the form

y^2           x^2

___    -    _____    =  1

 16             9

Here's the graph :  https://www.desmos.com/calculator/yq7wtlrut2

Parabola

Formally, a parabola is defined as follows: For a given point, called the focus, and a given line not through the focus, called the directrix, a parabola is the locus of points such that the distance to the focus equals the distance to the directrix.

Hyperbola

Formally, a hyperbola can be defined as follows: For two given points, the foci, a hyperbola is the locus of points such that the difference between the distances to each focus is constant.

The asypmtotes of a hyperbola will be perpendicular when a  = b

See here as an example :   https://www.desmos.com/calculator/lahqwjj2fk

\(\frac{2}{3} t - 1 < t + 7 \le -2t + 15 \)

We first have  the inequality

2/3 t - 1 < t + 7    subtract 7 from both sides

2/3 t  -  8 <  t     subtract 2/3 t  from both sides

-8 < 1/3 t     multiply both sides by 3

-24 <  t

And we have the second inequalty

t + 7 ≤  -2t + 15       add 2t  to both sides...subtract 7 from both sides

3t ≤  8      divide both sides by 3

t ≤ 8/3

So      (-24, 8/3 ]    is the solution

See the pic, here :

Since ABC = 49  and ACB  = 12

Then BAC  =  180  - 49 - 12 =  119

Then  we have quadrilateral HFAE  

Angle HFA  = 90

Angle FAE  =  Angle BAC = 119

Angle AEH  = 90

So  angle FHE  = Angle BHC  =   360 - 90  - 90 - 119   =  61°

Let the weight of a sack of sugar  =  x

Let the weight of a sack of flour  = y

We have these inequalities

3x + 4y ≤  50

2y - 3x ≤ 13

And we wish to maximize y

Look at the graph, here :

https://www.desmos.com/calculator/aiwomrcer1

The  value that maximizes y, the weight of a sack of flour,  is   10.5 lbs

OK

So far  your team has played  20 games.......

And the number you have won =  3

So.....let x  be the number of additional weeks....since you win 3 games a week...you will win 3x additional games....and you will play 4x additional games

[ 3 + 3x]

_______   =  .65     cross-multiply

[ 20 + 4x]

3 + 3x  =  .65 [ 20 + 4x]      simplify

3 + 3x  =  13 + 2.6x      subtract 3, 2.6x  from both sides

.4x  =  10      divide both sides by .4

x  =  25   additional weeks..so...since you have already played 5 weeks...you need to play

25  +  5   =  

30 total weeks to get to a winning percentage of 65%

Proof

[3 + 3(25)]                78

_________   =      ______   =  .65

 20 + 4(25)               120

A) You can use the same diagram.....just add segment HE

This segment will form one leg of  right triangle HEJ   = 3960 miles

And JE forms the hypotenuse  = GE + JG =   3960 + 6  =  3966 miles

B)   So....using the Pythagorean Theorem

JH  = 

sqrt ( JE^2 - HE^2 )  =

sqrt  (3966^2 - 3960^2 )   =

sqrt (47556)  ≈

C)   218.1 miles = distance to the horizon 

We can use some symmetry to solve this

Draw a segment from the center of the square to the top right vertex of the square...this will divide the shaded area into to 2 equal parts...let this segment  be EC

If we let the center of the the circle on the right be the midpoint of BC  ....call this point, F....and the radius of this circle will  = 4

Then.....  the sector FEC of the circle on the right   will be the quarter area of this circle  = 

(1/4)pi (4)^2  =  4pi  units^2

And.....  the triangle formed by FEC  wll have the area  (1/2)(4)^2sin 90  = 8 units^2

So the top half of the shaded area is just the difference of these two areas  =

[ 4pi  -  8]    units^2

So....the whole shaded area is just twice this  =

[ 8pi - 16 ] cm^2  ≈   9.13 cm^2

Here's a pic, tertre :

A  = 30.5°

c  =  52.8

Angle C  =  90 - 30. 5   =  59.5°

Using the Law of Sines

a / sin A  =  c / sin C

a / sin 30.5 =    52.8 / sin 59.5          multiply both sides by sin 30.5

a =  52.8*sin 30.5  / sin 59.5    =  31.1 ft