Yes it will, 6cm and 17.5
All estimating problems make the assumption you are familar with your math facts, addition and multiplication. Since students normally memorize multiplication facts for single-digit numbers, any problem that can be simplified to single-digit numbers is easily worked.
2. You are asked to estimate 47.99 times 0.6. The problem statement suggests you do this by multiplying 50 times 0.6. That product is the same as 5 × 6, which is a math fact you have memorized. You know this because
.. 50 × 0.6 = (5 × 10) × (6 × 1/10)
.. = (5 × 6) × (10 ×1/10) . . . . . . . . . . . by the associative property of multiplication
.. = 30 × 1
.. = 30
3. You have not provided any clue as to the procedure reviewed in the lesson. Using a calculator,
.. 47.99 × 0.6 = 28.79 . . . . . . rounded to cents
4. You have to decide if knowing the price is near $30 is sufficient information, or whether you need to know it is precisely $28.79. In my opinion, knowing it is near $30 is good enough, unless I'm having to count pennies for any of several possible reasons.
See the image below, hope it helps!
Answer:
R = sqrt[(IWL)^2/(E^2 - I^2)] or R = -sqrt[(IWL)^2/(E^2 - I^2)]
Step-by-step explanation:
Squaring both sides of equation:
I^2 = (ER)^2/(R^2 + (WL)^2)
<=>(ER)^2 = (I^2)*(R^2 + (WL)^2)
<=>(ER)^2 - (IR)^2 = (IWL)^2
<=> R^2(E^2 - I^2) = (IWL)^2
<=> R^2 = (IWL)^2/(E^2 - I^2)
<=> R = sqrt[(IWL)^2/(E^2 - I^2)] or R = -sqrt[(IWL)^2/(E^2 - I^2)]
Hope this helps!
I think it's need to be a graph right?
