15925/37
Is the answer to your question
(-2,5)(1,1)
slope = (y2 - y1) / (x2 - x1)
slope = (1-5) / (1 - (-2) = -4/3
y = mx + b
slope (-4/3)
(1,1)..x = 1 and y = 1
now we sub
1 = -4/3(1) + b
1 = -4/3 + b
1 + 4/3 = b
3/3 + 4/3 = b
7/3 = b
y = -4/3x + 7/3.....(0,T)...x = 0
y = -4/3(0) + 7/3
y = 7/3
check...
(1,1)(0,7/3)
slope = (7/3 - 1) / (0 - 1)
slope = (7/3 - 3/3) / -1
slope = (4/3)/-1 = -4/3
if I did this correctly, I am thinking T = 7/3
Y=-5 would be a horizontal line going through -5
Answer:
37. {-1, -1}.
Step-by-step explanation:
I'll solve the first one . The other can be solved in a similar way. We can use the method of elimination.
x1 - x2 = 0
3x1 - 2x2 = -1
We can multiply the first equation by -2. We then have an equation containing + 2x2 so when we add this to the second equation the 2x2 will be eliminated
So the first equation becomes:
-2x1 + 2x2 = 0 Bring down the second equation:
3x1 - 2x2 = -1 Now adding, we get:
x1 + 0 = -1
so x1 = -1.
Now we substitute this value of x1 in the original first equation:
-1 - x2 = 0
-1 = x2
x2 = -1.
So the solution set is {-1, -1}.
If there are more than 2 equations you can use a combination of substitutions and eliminations.
Just has to be greater than or equal to 0... w and t can never be negative since taking the square root of a negative number yields an imaginary number which would mean imaginary current. Also the question states that w can't be zero but if T is zero than the current will be zero.
