Assuming there are two additional width-sized fences to make the division, then
2L + 4W = 1200 ie L + 2W = 600.
There are many possible dimensions: L 500, W 50; L 400, W 100 etc etc
Graph
Placing a question mark at the end of a phrase does not make it a sensible question. Try to use a whole sentence to describe what it is that you want answered.
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Placing a question mark at the end of a list of expressions or numbers does not make it a sensible question. Try to use a whole phrase (sentence) to describe what it is that you want answered.
It is an algebraic expression, if there are numbers and variables. For example: 2x + 2 = 4x(7-5) 20-29 = -9/3x If there are ONLY numbers and NO variables, it is called an arithmetic expression. For example: 3+3+3+3 = 4+4+4 2+6 = 4+4
a veterinarian uses 600 feet of chain-link fence to enclose a rectangular region. It is subdivided evenly into two smaller rectangles by placing fence parallel to one of the sides. what is the width (w) as a function of the length (L)? (w=?) what is the total area as a function of (L)? (A=?) what are the dimensions that produce the greatest enclosed area? EXPLENATION PLEASE
install a covering to enclose
Cumlitive output
This is to enclose the horse and the device is either called a lug or a tool bar. This is to enclose the horse and the device is called a lug.
by placing 5 cubes in one row and another in above row.
Continuity, resistance?
Placing another resistor in parallel to an existing resistor will lower the total resistance in the circuit. RParallel = 1 / Summationi=1toN (1 / Ri)
When batteries are connected in parallel, the voltage remains the same as the voltage of a single battery. This is because the positive terminals are connected together and the negative terminals are connected together, so the voltage across each battery remains constant.
It depends on what is wrong with the resistor. If it is damaged, replace it with the same type, value and power rating. If it is the wrong value and is not damaged, you can increase the value by placing another in series or reduce the value by placing another in parallel. If the resistor is variable type, you should be able to adjust it.
An ammeter's coil requires very little current for full-scale deflection (fsd). So, to measure a current above its fsd value, most of that current must be allowed to bypass the coil. This is achieved by placing a very low value shunt resistance in parallel with the coil ('shunt' is an archaic word for 'parallel').
No. The capacitor may improve its power factor by altering the effective reactive power of the machine, thus reducing the current drawn from the supply.
A: big difference diodes are semiconductor with the right polarity it can be a short to the power supply reversing will cause no problem and probably eliminate some spikes