The load that can be supported by a rectangular beam there is jointly as the width of the beam in the square of its height and inversely as the length of the beam a been 14 feet long with the width of 4 inches and the height of 6 inches can support a maximum load of 800 pounds if a similar board has a width of 9 inches and the height of 5 inches how long must be to support 1400 pounds

  12.5 feet

Step-by-step explanation:

The description of the load-carrying capacity of the beam translates as ...

  load = k·w·h²/l

Filling in the numbers, we can find the constant k:

  800 lb = k·(4 in)(6 in)²/(14 ft)

  k = (800·14 lb·ft)/(144 in³) = 700/9 lb·ft/in³

__

Then the length to support 1400 lb with the other given dimensions will be ...

  1400 lb = (700/9) lb·ft/in³ × (9 in)(5 in)²/l

  l = 700·25/1400 ft = 12.5 ft

The beam must be at most 12.5 ft long to support 1400 pounds.

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!