HI can someone do these 5 problems? it will really help me out I'll give out brainliest to the person who answers them. (No links) ​

$73.60

$345

simple interest = amount deposited x time x interest rate

600 + (600 x 0.055 x 5) = $765

600 + (600 x 0.055 x 5) > $2000

$765  $2000

He would not have $2000 in 5 years

Step-by-step explanation:

Total cost of items purchased = $75 + (2 x $8.50) = 92

If there is a 20% discount, he would pay (100 - 20%) 80% of the total cost =

0.8 x $92 = $73.60

commission earned = percentage commission x amount of sales

10% x $3450

= 0.1 x 3450 = 345

Amount he would have in his account = amount deposited + simple interest

simple interest = amount deposited x time x interest rate

600 x 0.055 x 5 = $165

Amount in his account in 5 years = $165 + 600 = $765

He would have less than $2000 in his account. he would have $765

  40 minutes

Step-by-step explanation:

The circumference of the larger circle is ...

  C = 2πr = 2π(5 in) = 10π in

The bug navigates the circumference at 3π in/min, so will take

  time = distance/speed = (10π in)/(3π in/min) = 10/3 min

to travel once around.

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The circumference of the smaller circle is ...

  C = 2πr = 2π(2 in) = 4π in

The bug navigates this circumference at 2.5π in/min, so will take

  (4π in)/(2.5π in/min) = 8/5 min

to travel once around.

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The bugs will meet at a time that is the least common multiple of these times. Both can be expressed in 15ths of a minute as ...

  {50/15, 24/15}

Then the LCM of these will be ...

  (1/15)LCM(50, 24) = (1/15)(50×24)/GCD(50, 24) = 1200/30 = 40

It will be 40 minutes before the bugs next meet at point P.

___

A graphing calculator can make use of the mod function to show when the bugs meet at point P (total is displacement of the two bugs from P is zero). It shows the meeting occurs after 40 minutes.