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leva [86]
3 years ago
10

Aidan drives to school and back each day. The school is 16 miles from his home. He averages 40 miles per hour on his way to scho

ol. If his total trip takes 1 hour, at approximately what speed does Aidan drive home?
Mathematics
1 answer:
Fed [463]3 years ago
6 0
First we have to answers stand that speed is equivalent to the distance traveled in a certain time span. That is speed = distance/time. Conversely, time = distance/speed. We are asked to find the speed going home given, distance, total time and speed going to school. We can answer this equation by adding the time Aidan is spending going and coming back from school 

t = (distance to school)/(speed to school) + (distance to home)/(speed to home)

1hour = (16)/(40) + (16)/(speed to home) 

Speed to home = 26.67miles/hour 
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Equivalent expressions are expressions with same simplified form. Equivalent expressions for the given expression are;

  • Expression 2:  \ln(a) + 2\ln(b) = \ln(a) + \ln(b^2)  = \ln(ab^2)
  • Expression 3:  \ln(a^2) + \ln(b^2) - \ln(a) = \ln(a^2b^2/a) = \ln(ab^2)
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<h3>What are equivalent expressions?</h3>

Those expressions who might look different but their simplified forms are same expressions are called equivalent expressions.

To derive equivalent expressions of some expression, we can either make it look more complex or simple. Usually, we simplify it.

<h3>What is logarithm and some of its useful properties?</h3>

When you raise a number with an exponent, there comes a result.

Lets say you get

a^b = c

Then, you can write 'b' in terms of 'a' and 'c' using logarithm as follows

b = log_a(c)

Some properties of logarithm are:

log_a(b) = log_a(c) \implies b = c\\\\\log_a(b) + log_a(c) = log_a(b \times c)\\\\log_a(b) - log_a(c) = log_a(\frac{b}{c})\\\\log_a(b^c) = c \times log_a(b)\\\\log_b(b) = 1\\\\ log_a(b) + log_b(c) = log_a(c)

Log with base e = 2.71828.... is written as \ln(x) simply.

The expression given is 2\ln(a) + 2\ln(b) - \ln(a)

We get its simplified form as

2\ln(a) + 2\ln(b) - \ln(a) = \ln(a) + \ln(b^2) = \ln(ab^2)

Simplifying given expressions:

  • Expression 1:  \ln(ab^2) - \ln(a) = \ln(ab^2/a)  = \ln(b^2)

This isn't same as simplified form of original

. Thus , this expression is not equivalent to the given expression.

  • Expression 2:  \ln(a) + 2\ln(b) = \ln(a) + \ln(b^2)  = \ln(ab^2)

This is same as simplified form of original expression. Thus , this expression is equivalent to the given expression.

  • Expression 3:  \ln(a^2) + \ln(b^2) - \ln(a) = \ln(a^2b^2/a) = \ln(ab^2)

This is same as simplified form of original expression. Thus , this expression is equivalent to the given expression.

  • Expression 4:  2\ln(ab) = \ln((ab)^2) = \ln(a^2b^2)

This isn't same as simplified form of original expression. Thus , this expression is not equivalent to the given expression.

  • Expression 5:  \ln(ab^2)

This is same as simplified form of original expression. Thus , this expression is equivalent to the given expression.

Thus, equivalent expressions for the given expression are;

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  • Expression 5:  \ln(ab^2)

Learn more about equivalent expressions here:

brainly.com/question/10628562

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