Answer: Option <span>C. To get system B, the second equation in system A was replaced by the sum of that equation and the first equation multiplied by -3. The solution to system B will be the same as the solution to system A.
First equation of system A multiplied by -3:
(-3)(x+6y=5)
(-3)(x)+(-3)(6)=(-3)(5)
-3x-18=-15
Sum of the second equation of system A and the first equation multiplied by -3:
(-3x-18)+(3x-7y)=(-15)+(-35)
-3x-18+3x-7y=-15-35
-25y=-50
System B
x+6y=5
-25y=-50 </span>
Answer:
Class 4: This class of truck has a GVWR of 14,001–16,000 pounds or 6,351–7,257 kilograms. Class 5: This class of truck has a GVWR of 16,001–19,500 pounds or 7,258–8,845 kilograms.
Step-by-step explanation:
Answer:
<em>The voltage at the middle source is</em> 
Step-by-step explanation:
<u>Voltage Sources in Series</u>
When two or more voltage sources are connected in series, the total voltage is the sum of the individual voltages of each source.
The figure shown has three voltage sources of values:



The sum of these voltages is:

Operating:

We know the total voltage is
, thus:

Equating the real parts and the imaginary parts independently:
4+a=6
1+b=-3
Solving each equation:
a = 2
b = -4
The voltage at the middle source is 
Answer:
b. 7/18.
Step-by-step explanation:
That would be 1/2 - 1/9
= 9/18 - 2/18
= 7/18.
Conjugate of 9-5i is 9+5i
product of those two: (9-5i)*(9+5i) =