It is known that a cable with a cross-sectional area of 0.300.30 sq in. has a capacity to hold 2500 lb. If the capacity of the cable is proportional to its cross-sectional area, what size cable is needed to hold 70007000 lb?

0.84 square in.

Step-by-step explanation:

Since the capacity of the cable is proportional to its cross-sectional area.

Mathematically,

C = k * A

A1 = 0.3 sq

C1 = 2500 lb

C2 = 7000 lb

C1/A1 = C2/A2

2500/0.3 = 7000/C2

= 7000 / 8333.33

= 0.84 square in.

5 x 14?

im new, sorry if this is wrong.

step-by-step explanation:

the answer is b: 18

step-by-step explanation:

use the pythagorean theorem to .

a^2+b^2=c^2

plug 18 for "r"

the 2 legs are 18 and 24

18^2+24^2=c^2

324+576=c^2

900=c^2

c=30

30-12=18

18 is the radius

: )