Answer:
56 different tests
Step-by-step explanation:
Given:
Number of wires available (n) = 8
Number of wires taken at a time for testing (r) = 5
In order to find the number of different tests required for every possible pairing of five wires, we need to find the combination rather than their permutation as order of wires doesn't disturb the testing.
So, finding the combination of 5 pairs of wires from a total of 8 wires is given as:
Plug in the given values and solve. This gives,
Therefore, 56 different tests are required for every possible pairing of five wires.
Answer:
Step-by-step explanation:
There are several ways to solve this quartic equation. But since the coefficients, they repeat a=1,b=2,c=1,d=2, but e=0, and they are multiple of each other, then it is more convenient to work with factoring as the method of solving it.
As if it was a quadratic one.
![x^{4}-2x^{3}- x^{2} + 2x = 0\\x(x^{3}-2x^{2}-x+2)=0 \:Factoring \:out\\x[(x^{3}-2x^{2})+(-x+2)]=0 \:Grouping\\x[\mathbf{x^{2}}(x-2)+\mathbf{-1}(x+2)]=0 \:Rewriting\:the\:first\:factor\\x(x^{2}-1)(x-2)\:Expanding \:the \:first \:factor\\x(x-1)(x+1)(x-2)=0\\x=0,x=1,x=-1,x=2\\S=\left \{ 0,-1,1,2 \right \}](https://tex.z-dn.net/?f=x%5E%7B4%7D-2x%5E%7B3%7D-%20x%5E%7B2%7D%20%2B%202x%20%3D%200%5C%5Cx%28x%5E%7B3%7D-2x%5E%7B2%7D-x%2B2%29%3D0%20%5C%3AFactoring%20%5C%3Aout%5C%5Cx%5B%28x%5E%7B3%7D-2x%5E%7B2%7D%29%2B%28-x%2B2%29%5D%3D0%20%5C%3AGrouping%5C%5Cx%5B%5Cmathbf%7Bx%5E%7B2%7D%7D%28x-2%29%2B%5Cmathbf%7B-1%7D%28x%2B2%29%5D%3D0%20%5C%3ARewriting%5C%3Athe%5C%3Afirst%5C%3Afactor%5C%5Cx%28x%5E%7B2%7D-1%29%28x-2%29%5C%3AExpanding%20%5C%3Athe%20%5C%3Afirst%20%5C%3Afactor%5C%5Cx%28x-1%29%28x%2B1%29%28x-2%29%3D0%5C%5Cx%3D0%2Cx%3D1%2Cx%3D-1%2Cx%3D2%5C%5CS%3D%5Cleft%20%5C%7B%200%2C-1%2C1%2C2%20%5Cright%20%5C%7D)
Answer:
a) How much work has he done = dW = 1.3 × 105 J
b) What is the magnitude of the heat that he has lost = dQ = -5.4 × 105 J
Step-by-step explanation:
- From the first law of thermodynamics; dQ = dU + dW
Hence, work done dW = dQ - dU
dW = -6.7 × 105 J - ( -8.0 × 105 J)
dW = 1.3 × 105 J is the work done by the player
- for heat lost; from dQ = dW + dU
- dU = -8.0 × 105 J and dW = 2.6 × 105 J
- heat lost dQ = 2.6 × 105 J + ( -8.0 × 105 J)
- The negative is as a result of heat lost by the player
