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KiRa [710]
2 years ago
11

Adrian is planning to drive from City X to City Y. The scale drawing below shows the distance between the two cities with a scal

e of ¼ inch = 18 miles.
City X to City Y is 3 1/2 in.
If Adrian drives at an average speed of 30 miles per hour during the entire trip, how much time, in hours and minutes, will it take him to drive from City X to City Y?
Mathematics
1 answer:
DochEvi [55]2 years ago
4 0

Answer:

takes him 8.4 hrs

Step-by-step explanation:

if 1/4=18inch then 3 1/2=7/2 so 7/2 divide by 1/4=14 then 18 x 14=252

so distance is 252 miles

speed is 30mph so 252/30=8.4h

the answer might need the 0.4 in hrs too

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Drag the tiles to the correct boxes to complete the pairs.
Len [333]

<u>Answer:</u>

1. (-2^2)^{-6} ÷ (2^{-5})^{-4} \implies 2^{-32}

2. 2^4 . (2^2)^{-2} \implies 1

3. (-2^{-4}).(2^2)^0 \implies -2^8

4. (2^2).(2^3)^{-3} \implies 2^{-5}

<u>Step-by-step explanation:</u>

1. (-2^2)^{-6} ÷ (2^{-5})^{-4} :

= \frac{ ( - 2 ^ 2 ) ^ { - 6 } } { ( 2 ^ { - 5 } ) ^ { - 4 } } = \frac{2^{-12}}{2^{20}} = 2^{-12-20}=2^{-32}

2. 2 ^ 4 . ( 2 ^ 2 ) ^ { - 2 } :

= 2^4 \times \frac{1}{2^4} = 1

3. (-2^{-4}).(2^2)^0 :

= (-2^4)^2 \times 1 = -2^8

4. (2^2).(2^3)^{-3} :

= 2^4 \times \frac{1}{2^9} =\frac{1}{2^5} =2^{-5}

5 0
3 years ago
Read 2 more answers
14. Find the coordinates of the circumcenter for ∆DEF with coordinates D(1,1) E (7,1) and F(1,5). Show your work.
Bess [88]
Hello : 
 the  <span>coordinates </span> circumcenter for ∆DEF : 
x = (1+7+1)/3
y = (1+1+5//3
x=3
y= 7/3
6 0
3 years ago
Write a word problem that would have the solution <br> p&lt;21<br> Lineunder neath the
Dimas [21]
The line underneath the inequality line(arrow) means p is less than OR EQUAL to 21.
An example off the top of my head would be, Rachel just started a new job that pays 7 dollars per hour. She only worked 3 hours and decided she wanted to go to the mall to buy some clothes. How much money does Rachel have to spend?
And your answer is right there
5 0
3 years ago
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JulsSmile [24]
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5 0
2 years ago
How to solve an expression with variables in the exponents?
gtnhenbr [62]

Answer:

  use logarithms

Step-by-step explanation:

Taking the logarithm of an expression with a variable in the exponent makes the exponent become a coefficient of the logarithm of the base.

__

You will note that this approach works well enough for ...

  a^(x+3) = b^(x-6) . . . . . . . . . . . variables in the exponents

  (x+3)log(a) = (x-6)log(b) . . . . . a linear equation after taking logs

but doesn't do anything to help you solve ...

  x +3 = b^(x -6)

There is no algebraic way to solve equations that are a mix of polynomial and exponential functions.

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Some functions have been defined to help in certain situations. For example, the "product log" function (or its inverse) can be used to solve a certain class of equations with variables in the exponent. However, these functions and their use are not normally studied in algebra courses.

In any event, I find a graphing calculator to be an extremely useful tool for solving exponential equations.

8 0
2 years ago
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