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.

i think it's either c or d

step-by-step explanation: the entire question is extremely odd. 04 obviously refers to the interest rate. honestly i will say the answer is c. because the (2 x 12) part would mean it's compounded semiannually. i'll say choose c.

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0