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Digiron [165]
4 years ago
15

Susan drove 1500 miles to Daytona Beach for spring break.On the way back she averaged 10 miles per hour less, and the drive back

took her 5 hours longer . Find Susan's average speed on the way to Daytona Beach.
Mathematics
1 answer:
Nataly_w [17]4 years ago
7 0
Hey im in a corrections faculity would you write me plz

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Please help me with my homework!!!!
Zina [86]

Answer:

∠ACB= 57°

AB= 8.80 units (3 s.f.)

AC= 10.1 units (3 s.f.)

Step-by-step explanation:

Please see attached picture for full solution.

8 0
3 years ago
Describe a relationship in which it would be reasonable to have a negative number in the domain or
Aliun [14]

Answer:

Relationship between temperature at different times of the day during the Winter period in places like Alaska and Minnesota in the USA.

Step-by-step explanation:

An example of a relationship that would have a negative number in the domain or range would be the relationship between the temperature at different times of the day during the snow period in the winter. It's well known that in places like Alaska and Minnesota in the USA, that during the winter when there's very heavy snow, the average temperature is usually below 0°C. The temperature is the dependent variable and is most likely negative in those 2 states and thus the range would have negative numbers.

7 0
3 years ago
What is the interquartile range of this data set?<br> 2,5,9,11,18, 30, 42, 48, 55, 73, 81
Lubov Fominskaja [6]

Answer:

46

Step-by-step explanation:

55-9=46

2=min

9=Lq

30=median

55= Uq

81= max

3 0
4 years ago
How do I do this problem
Marina86 [1]
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation : 

                    <span> 5*(x-2)-4*(x+1)-(50)=0 </span>
4 0
4 years ago
Solve using long division <br> Please
madreJ [45]

1. Solution,\frac{2x^3+4x^2-5}{x+3}:\quad 2x^2-2x+6-\frac{23}{x+3}

Steps:

\mathrm{Divide}\:\frac{2x^3+4x^2-5}{x+3}:\quad \frac{2x^3+4x^2-5}{x+3}=2x^2+\frac{-2x^2-5}{x+3}

\mathrm{Divide}\:\frac{-2x^2-5}{x+3}:\quad \frac{-2x^2-5}{x+3}=-2x+\frac{6x-5}{x+3}

\mathrm{Divide}\:\frac{6x-5}{x+3}:\quad \frac{6x-5}{x+3}=6+\frac{-23}{x+3}

\mathrm{Simplify}, =2x^2-2x+6-\frac{23}{x+3}

\mathrm{The\:Correct\:Answer\:is\:2x^2-2x+6-\frac{23}{x+3}}

2. Solution, \frac{4x^3-2x^2-3}{2x^2-1}:\quad 2x-1+\frac{2x-4}{2x^2-1}

Steps:

\mathrm{Divide}\:\frac{4x^3-2x^2-3}{2x^2-1}:\quad \frac{4x^3-2x^2-3}{2x^2-1}=2x+\frac{-2x^2+2x-3}{2x^2-1}

\mathrm{Divide}\:\frac{-2x^2+2x-3}{2x^2-1}:\quad \frac{-2x^2+2x-3}{2x^2-1}=-1+\frac{2x-4}{2x^2-1}

\mathrm{The\:Correct\:Answer\:is\:2x-1+\frac{2x-4}{2x^2-1}}

\mathrm{Hope\:This\:Helps!!!}

\mathrm{-Austint1414}

4 0
3 years ago
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