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vladimir1956 [14]
2 years ago
13

Suppose that a bored student wants to go to a restaurant for lunch, but she only has an hour in which to go, eat, and get back i

n time for class. Considering that it usually takes about 30 minutes in most restaurants to get served and to eat, what is the farthest restaurant the student can go to without being late for class? assume in this part that the student has a car that can accelerate to its top speed in a negligible amount of time. Also, the local speed limit is 30 mph and the student would not like to get a speeding ticket.
SAT
1 answer:
Bas_tet [7]2 years ago
8 0

Well, it really depends, one would need to know the restraunts around them and know how long it would to take to get to them.

Now lets say there is one 10 minutes away from the school, but the serving time is a about 30 minutes, then again it would also depend on how fast the student can eat, assuming it takes about 20-30 minutes, the student would most likely be 10-20 minutes late.

Hope this helped!! (And good luck with getting the food!!) Have a nice day!!

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Factor the expression into an equivalent form 12y^2-75.
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Hello oddworld7836!

\huge \boxed{\mathbb{QUESTION} \downarrow}

Factor the expression into an equivalent form 12y² - 75.

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

12 y ^ { 2 } - 75

By observing the expression, we can see that, 3 is the only common factor in both the terms of the expression. So, take the common factor 3 out.

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