w2 - 13w + 48
You want to write the equation in the form (w+a)(w+b). However, if this particular equation could be written in this form, there would be at least one value of x where y = 0. There are no such solutions for this equation, so it cannot be factored into this form.
However, you can factor it if you separate 48 into 42.25 + 5.75:
w2 - 13w + 48 = (w - 6.5)2 + 5.75 = (w - 6.5)(w - 6.5) + 5.75
(2w + 3)(w + 5)
(x+5)(2x+3)
13 multiplied by the amount (quantity) 60 minus w is written mathematically as 13(60-w).
-13w
HELICAL 13W 120VAC 60Hz 180mA13W means 13 watts cfl...... =40 watt incadescent.
There is no equation, just an expression. And, since it is an expression, it cannot be solved so w can equal any value whatsoever.
Longstar FE-IISB-13W 27K
A = wl91 = w1391/13 = w13/13w = 7m
I tried to twist it in several directions but it didn't work.
grandet quwen 1i4o2eru biuer n 13w qedv `47v q3wev`3e7 ` 134ev q23e `2344444444444444
Yes, a 15W CFL lamp can generally be substituted for a 13W CFL lamp, as both are designed to fit standard sockets and operate on similar voltage. However, the 15W lamp will produce more light (lumens) and may consume slightly more energy. Ensure that the fixture can accommodate the higher wattage to avoid overheating or damage. Always check the manufacturer's specifications for compatibility.
Fluorescents are always 3-4 times more efficient than halogens which are a version of incandescent bulb.