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Alex73 [517]
3 years ago
7

Mitchell ran a total of 32 miles over the course of 16 track practices. How many miles would Mitchell have run after 23 track pr

actices?
Mathematics
2 answers:
zhuklara [117]3 years ago
7 0

Given:

Mitchell ran a total of 32 miles over the course of 16 track practices.

To find:

How many miles would Mitchell have run after 23 track practices?

Solution:

We have,

Total miles over the course of 16 track practices = 32 miles

Total miles over the course of 1 track practices =  miles

                                                                             = 2 miles

Total miles over the course of 23 track practices =  miles

                                                                                = 46 miles

Therefore, Mitchell would run 46 miles.

Allushta [10]3 years ago
3 0

Given:

Mitchell ran a total of 32 miles over the course of 16 track practices.

To find:

How many miles would Mitchell have run after 23 track practices?

Solution:

We have,

Total miles over the course of 16 track practices = 32 miles

Total miles over the course of 1 track practices = \dfrac{32}{16} miles

                                                                              = 2 miles

Total miles over the course of 23 track practices = 23 \times 2 miles

                                                                                 = 46 miles

Therefore, Mitchell would run 46 miles.

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3 years ago
What is the distance between the points (5.-18) and (8,
romanna [79]

Answer:

\sqrt{10}

Step-by-step explanation:

The horizontal distance from points (5,-18) and (8,-17) is 3 because it is 3 units from 5 to 8.  The vertical distance is 1 since it is one unit from -18 to -17.  Now we can use the equation a^{2}+b^{2}=c^{2} where a=3 and b=1 and c is the distance that you are looking for:

a^{2} +b^{2} =c^{2}\\3^{2} +1^{2} =c^{2}\\9+1=c^{2}\\10=c^{2}\\\sqrt{10}=c

8 0
3 years ago
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6 0
3 years ago
What is the sum of 20x^2-10x-30
ivann1987 [24]

Answer:

The sum of the roots is 0.5

Step-by-step explanation:

<u><em>The correct question is</em></u>

What is the sum of the roots of 20x^2-10x-30

we know that

In a quadratic equation of the form

ax^{2} +bx+c=0

The sum of the roots is equal to

-\frac{b} {a}

in this problem we have

20x^{2} -10x-30=0  

so

a=20\\b=-10\\c=-30

substitute

-\frac{(-10)} {20}=0.5

<u><em>Verify</em></u>

Find the roots of the quadratic equation

The formula to solve a quadratic equation is equal to

x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}

a=20\\b=-10\\c=-30

substitute

x=\frac{-(-10)\pm\sqrt{-10^{2}-4(20)(-30)}} {2(20)}

x=\frac{10\pm\sqrt{2,500}} {40}

x=\frac{10\pm50} {40}

x=\frac{10+50} {40}=1.5

x=\frac{10-50} {40}=-1

The roots are x=-1 and x=1.5

The sum of the roots are

-1+1.5=0.5 ----> is ok

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