Make t the subject of the familiar:
3π
(ππ‘βπ)
= 1 + U
A. t =
3π
(1+ π’}
+
π
π
B. t =
3π
π£(1+ π’}
+
π
π
C. t =
3
(1+ π’}
+
π
π
D. 3
π£(1+ π’}
+
π
π
To make t the subject of the equation, we want to isolate t on one side of the equation.
Starting with the given equation:
(3π) / (ππ‘ β π) = 1 + U
We can start by multiplying both sides of the equation by (ππ‘ β π) to eliminate the denominator:
3π = (ππ‘ β π) * (1 + U)
Next, we distribute the (1 + U) term on the right side of the equation:
3π = ππ‘(1 + U) β π(1 + U)
Expanding the equation further:
3π = ππ‘ + ππ‘U β π β πU
Next, we want to gather all the terms containing π‘ on one side of the equation. We do this by subtracting ππ‘ from both sides:
3π - ππ‘ = ππ‘U β π β πU
Rearranging the terms:
π‘(3π β π) = ππ‘U β π β πU
Finally, we want to isolate t, so we divide both sides of the equation by (3π β π):
π‘ = (ππ‘U β π β πU) / (3π β π)
The final result is:
π‘ = (ππ‘U β π β πU) / (3π β π)
Therefore, the correct option is D. 3π£(1+ π’)/(π£ - π)
To make t the subject of the equation 3π / (ππ‘βπ) = 1 + U, we follow these steps:
Step 1: Multiply both sides of the equation by (ππ‘βπ):
(ππ‘βπ) * (3π / (ππ‘βπ)) = (1 + U) * (ππ‘βπ)
Simplifying the left-hand side:
3π = (1 + U) * (ππ‘βπ)
Step 2: Expand the right-hand side:
3π = (1 + U)ππ‘β(1 + U)π
Step 3: Distribute (1 + U) to terms inside the parentheses:
3π = ππ‘ + πππ‘ β π β ππ
Step 4: Combine like terms:
3π = ππ‘ + πππ‘ β π β ππ
3π = (1 + π)ππ‘ β (1 + π)π
3π = (1 + π)ππ‘ β π(1 + π)
Step 5: Move π(1 + π) to the right-hand side:
3π + π(1 + π) = (1 + π)ππ‘
Step 6: Divide both sides of the equation by (1 + U)π:
(3π + π(1 + π)) / (1 + π)π = π‘
Finally, we have:
π‘ = (3π + π(1 + π)) / [(1 + π)π]
Therefore, the answer is A. t = (3π + π(1 + π)) / [(1 + π)π].