Brian has $78 and wants to purchase tapes through a music club. Each tape costs $7.50. The music club will add a total postage and handling charge of $3.50 to his order. What is the greatest number of tapes he can purchase?

Based on the cost of the tape, the amount Brian has and the handling charge, Brian can purchase no more than 9 tapes.

First find out the amount he can spend on tapes:

= Total amount to use - Postage and handling charge

= 78 - 3.50

= $74.50

Each tape costs $7.50 so the highest number of tapes he can buy is:

= Amount available for tapes / Cost of tapes

= 74.50 / 7.50

= 9.93 tapes

= 9 tapes

Note: Round down to 9 because the tapes bought have to be a whole number and cannot be rounded up without more money.

In conclusion, Brian can only buy 9 tapes.

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9 tapes

Step-by-step explanation:

First, subtract the handling charge from his $78.

78 - 3.50 = 74.5

So, Brian has $74.50 to purchase tapes. Each tape costs $7.50, so let's find out how many he can purchase.

74.50 / 7.50 = 9.933 (approximately)

Since Brian can't buy 0.933 of a tape, he will only be able to buy 9 tapes.

step-by-step explanation:

here's what happens:

1) the graph of f   = x^2 is shifted 2 units to the right.

2) the resulting graph is reflected in the x-axis, so that the parabola now opens down.

3) this final result is transformed up by 5 units.

hi there!

first we need to find out how much they are adding to each number to make the next number.

ex: 4 + 4 = 8, 8 + 8 = 16, and so on.

so continuing with 32,

32 + 32 = 64

64 + 64 = 128

128 + 128 = 256

256 + 256 = 512

4, 8, 16, 32, 64, 128, 256, 512

the 8th term of the sequence will be 512

hope this ! ~