(brainliest) plz i cant put a lot of point for this cause i dont got a lot but someone plz suppose a friend is having trouble with exponents and asks you to review some worked problems. look over the problems and correct any mistakes. explain the errors in the student’s reasoning.
)/(2a^6b^2) = 4/2 ×(a^2)/(a^6)×(b^5)/(b^2)
= 2 × a^4× b^3
2a^4b^3
b). (y^2)/(y^7)= y^-14
= 1/(y^14)
(12k^2m^3n)/(9m^3n^6k^5) =– 12/9×(k^2)/(k^5)×(m^3)/(m^3)×(n^6)/ n
=-4/3× k^(-3) ×m^3×n^5
=-(4m^3n^5)/(k^3)
d.) 2x^2y×3xy^5y^2=(2×3)(x^2× x^5)(y×y^2)
=6x^10y^2
)^3×2x^4=3x^8×2x^4
=6x^32

omg what happend to your question? if you could edit this question i can

step-by-step explanation:

relating to or denoting a system of numbers and arithmetic based on the number ten, tenth parts, and powers of ten.

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!