Answer:
<u>Solution</u><u> </u><u>given</u><u>:</u>
<u>one</u><u> </u><u>side</u><u> </u><u>of</u><u> </u><u>a</u><u> </u><u>triangle</u><u> </u><u>must</u><u> </u><u>be</u><u> </u><u>less</u><u> </u><u>than</u><u> </u><u>the</u><u> </u><u>sum</u><u> </u><u>of</u><u> </u><u>two</u><u> </u><u>other</u><u> </u><u>side</u><u> </u><u>.</u>
<u>by</u><u> </u><u>making</u><u> </u><u>these</u><u> </u><u>sense</u><u>:</u>
3<10+20
10<20+3
20<10+3not true
these three side are not possible.
again
10<20+25
20<25+10
25<20+10
these three side are true so
required side are:
Side 1:10
Side 2:25
Side 3:20
Answer:
The domains of both f(x) and g(x) must be (–∞, ∞).
Step-by-step explanation:
-Ash
:p
= 1/2kx2
U = potential energy of a spring at a certain position
k = the spring constant, specific to the spring, with units N/m.
x = distance the spring is stretched or compressed away from equilibrium
Potential Energy: Elastic Formula Questions:
1) A spring, which has a spring constant k = 7.50 N/m, has been stretched 0.40 m from its equilibrium position. What is the potential energy now stored in the spring?
Answer: The spring has been stretched x = 0.40 m from equilibrium. The potential energy can be found using the formula:
U = 1/2kx2
U = 1/2(7.50 N/m)(0.40 m)2
U = 0.60 N∙m
U = 0.60 J
Answer:
81 F
Step-by-step explanation:
Hours | Temp.
0 | 75F
1 | 77F
2 | 79F
3 | 81F