Answer:
(a) 85 - (35 x 2) - 6
Step-by-step explanation:
The miles Mr. Richardson has left to drive will be the difference between the distance to his goal and the miles he has alread driven. The relation between distance, speed, and time can be used to find the miles he drove before lunch.
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<h3>before lunch</h3>
distance = speed × time
miles before lunch = (35 mi/h) × (2 h) = (35 × 2) mi
<h3>total driven</h3>
The total miles driven before stopping for gas will be ...
miles before stopping for gas = (miles before lunch) + (6 more miles)
= 35 ×2 +6 . . . miles
<h3>miles remaining</h3>
Then the remaining miles are ...
remaining = trip miles - miles driven
= 85 -(35 ×2 +6)
= 85 - (35 ×2) -6 . . . . . miles left to drive
Y= -2(x)+7 I'm pretty sure....
Answer:
<h3>ans is 300</h3>
Step-by-step explanation:
<h3>you need to take LCM of 15,20,25 </h3><h3>and their LCM Will be 300</h3>
<h3>I'M NOT PRETTY SURE ABOUT IT. ...</h3>
Found this. Hope it helps.
https://www.algebra.com/algebra/homework/word/coins/Word_Problems_With_Coins.faq.question.1059726.html
The sale price would be 24.64
