Monday

March 2, 2015

March 2, 2015

Posted by **Karry** on Wednesday, May 1, 2013 at 9:07pm.

Parabola equation y= 4(100 - x^2/2500)

x and y are measured in feet

A. Find the length of the road across the bridge

B. find the height of the parabolic arch at the center of the span

C. Find the lengths of the vertical girders at intervals of 100 feet from the center of the bridge

( graph of upside down parabola)

problem is on photobucket

username: piggerockalbum

album: math problems

- college algebra -
**Reiny**, Wednesday, May 1, 2013 at 9:26pmI can't log into your photobucket, (needs password)

so I will "guess" at a diagram

expanding and simplifying your equation I got

y = (-1/625)x^ + 400

b) at the centre, x = 0

so y = 400

the bridge is 400 ft high

width of bridge, let y = 0 (on the x-axis)

0 = (-1/625)x^2 + 400

x^2 = (625)(400)

x = ±25(20) = ±500

So the bridge is 1000 ft wide

for vertical girders

let x = 100, 200, 300 , 400

x = 100

y = (-1/625)(100)^2 + 400 = 384

x = 200

y = (-1/625)(200)^2 + 400 = 336

etc.

**Answer this Question**

**Related Questions**

College Algebra - Find the standard form of the equation of the parabola with ...

College Algebra - Find the standard form of the equation of the parabola with ...

algebra word problem: still stuck - question: rachel allows herself 1 hour to ...

College Algebra - Theres this chemistry problem im getting stuck in because i'm ...

Integrating Factors - I've been working on this hw problem for a while now, but ...

Algebra story problem - I need to turn this story into an algebra problem: Your ...

Pre-Calc - Please help me with this problem-I'm stuck (x-2)/(x+2) = (x+10)/x I'm...

Biology - if the incidence of cystic fibrosis is 1 in 2500, what is the expected...

Honors Algebra II - I've been stuck on a problem with my homework for a couple ...

Physics - I am reposting this since I did so incorrectly the first time. (I ...