A light fixture at a traffic intersection operates on exactly 500 watts of power to light the bulbs in the fixture. Any combination of 150-watt, 100-watt, 75-watt, 60- watt or 50-watt bulbs can be used but the total number of watts must be 500. More than one of the same watt can be used. How many combinations could be used if at least one bulb must be 60 watt?

I found 5 combinations, key thing is that you need to have 5

60w bulbs

Step-by-step explanation: if we need to have at least one 60 W bulb, we notice

Five 60 W bulbs make 300W. Then we get missing 200 W with following combinations: 100 W + 100 W , 75 + 75 + 50 , 150 + 50 , 100+ 50 + 50

Or 4 times 50 W.

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

answer is 144

step-by-step explanation: