CPhill

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Circumference  =  2 * pi * r  = 2 * 3.14 * r  = 2 * 3.14 * 5.2   =  32.656 ft

y = -0.000488x^2  + 0.656x

We have the form  y  =  ax^2  + bx

The x coordinate of the vertex  is =  -b / [ 2 *a ] =      - 0.656  / [ 2 *  -0.000488 ] ≈  672.13

To find the  height above the river, put this value back into  the function and we  get

y  = -0.000488(672.13)^2  + 0.656(672.13) ≈ 220.5 ft

The section of the bridge above the arch will be the difference between  the intercepts of the function and the x axis

But  (0,0)  is one intercept...and using the fact that this parabola is symmetric bout its vertex point....the other  intercept occurs at   2 ( 672.13)  = 1344.26 ft ≈  1344.3 ft.

See the graph, here :  https://www.desmos.com/calculator/jd1toj6pak

[The line y = 220.5  represents  the bridge span  between the two  x intercepts  ]

The sum of the interior angles of a triangle  = 180°

Let  x  be the unknown angle...so we have...

55 + 80  +  x  = 180

135 + x  = 180     subtract  135 from both sides

x  = 45°

Force to to keep car from rolling down the hill... =   5300 sin (8°) ≈ 737.6 lbs

Force  perpendicular to the hill  =  5300 cos (8°) ≈  5248 4 lbs

 -2(s - 7) > 4s + 8   simplify the  left side

-2s + 14 > 4s + 8          add  2s to  both sides, subtract 8 from both sides

6 > 6s       divide both sides by 6

1 > s       which  means the same thing as     s < 1

So....the answer  is   ( - infinity, 1 )

If AK is an  altitude, then angle AKC  is   a right angle

By the Pythagorean Theorem...

AK  = √  [ AC^2  - CK^2 ] = √ [ 9^2  - 4^2 ] = √ [ 81 - 16] = √65

Also...angle  AKB is right....which implies that

AB  = √ [ AK^2 + BK^2 ] = √ [  (√ 65)^2 + (√5)^2 ] =  √ [65 + 5 ] =    √70  units^2

Triangle  ABC  is isosceles with  AB  = AC

AD  is an altitude which is also  a median to BC...so  DC  = DB  = 6

And  DE is parallel to AB

So  angle EDC  = angle ABC

And angle ACB  = angle ECD

So...by AA congruency, ΔABC  ~ Δ  EDC

Since the area  of  Δ  ABC = 180....we can  find the altitude AD thusly

180  = (1/2) (AC) (AD)

180  = (1/2) (12) (AD)

180 = 6  (AD)   divide each side by 6

30  = AD

Now  Δ ABC  and Δ EDC  are similar.... and BC  = 2DC

So......the altitude of Δ ABC  is also twice that of  Δ EDC...so  the altitude of Δ  EDC   = 15

And the area of Δ EDC  = (1/2)DC( altitude of EDC)  = (1/2)(6)(15)  = .45

So...the area  of ABDE  = area of  Δ ABC  -  area  of Δ  EDC  =   180  - 45  =  135  units^2

Let A  = (0,8)   B  =(0,0)   C  = (15,0)  and M  = (0,4)

So CM  = √ [ (15 - 0) ^2  + ( 0 - 4)^2 ] =   √ [15^2 + 4^2 ]   =  √ [ 225 + 16]  = √241 ≈ 15.52 units

Look  at "C'

If we multiply   25 * 1 hour  + 40   we get  1 * 25  + 40  = 25  + 40  = 65

If we multiply 2 * 2 hours + 40  we get 2 *25  + 40  = 50 + 40  = 90

This seems to  be the rule....check it yourself for 3 hours, 4 hours and 5 hours to see that this rule is correct....!!!