Aquadratic equation has the zeros -3 amd 6. can the quadratic equation be the given equation? a. (2x + 6)(x - 6) =0. yes or no b. (6x - 1)(x + 3) =0. yes or no c. -3x(x - 6) =0. yes or no

It can be A. (not B or C)

Step-by-step explanation:

It is A because x-6=0 can be simplified to x=6. Then, 2x + 6, you can divide the whole equation resulting in x+3=0, simplify this and you get x=-3. YES

It is not B because, while x+3=0 results in a zero of -3, 6x-1 can be simplified to be divided by 6. When we do this we get x-1/6=0, which is not equivalent to 6. NO

It is not C because, while x-6=0 results in a zero of 6, -3x can be simplified with the zero product property to get -3x=0 then dividing -3 by 0 giving you 0 which is not equivalent to one. NO

answer: sensy

step-by-step explanation:

if gloria has "carry over" minutes, (that is, any minutes paid for but not used in a month can be applied to her payment for minutes in the next month,) her total usage for the three months is: total minutes = 320 + 243 + 489 = 1052 minutes number of 200 minute blocks = 1052/200 = 5 r52 (that is 5 with a remainder of 52) rounding up the number of 200 minute blocks gives her usage for the three months as 6 blocks. the cost for six blocks is 6 x $25 = $150 she also has 200 - 52 or 148 carry over minutes if gloria pays each month and unused minutes are lost, applying the rounding up procedure to each month gives: first month usage of 320 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 second month usage of 243 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 third month usage of 489 minutes rounds up to 3 blocks of 200 minutes for a payment of 3 x $25 = $75 total payments for the three months are $50 + $50 + $75 = $175 get the plan with carry over minutes!