Tina recorded the number of calories she burned during a bike ride. The graph shows Tina’s calories burned, y, after x hours. If Tina rode her bike at a constant rate for the first three hours, what was Tina’s average calories burned, in calories per hour, between hour 0 and hour 3?

Answers

Answer 1

Answer:

the answer is B because 2x3 is ainzqwndwqncjnjqanc2ejn2qncb

Step-by-step explanation:

Answer 2

400 per hour was Tina's average calorie burn per hour between hours 0 and 3 if she rode her bike at a steady speed for the first three hours.

Given that,

Tina kept track of the calories she expended while riding her bike. After x hours, the graph displays Tina's calories burned, y.

We have to find what was Tina's average calorie burn per hour between hours 0 and 3 if she rode her bike at a steady speed for the first three hours.

What is a graph?

A graph is a visual representation or diagram that displays facts or values in an organized manner in mathematics. The relationships between two or more objects are often shown by the points on a graph.

Therefore, 400 per hour was Tina's average calorie burn per hour between hours 0 and 3 if she rode her bike at a steady speed for the first three hours.

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Tina Recorded The Number Of Calories She Burned During A Bike Ride. The Graph Shows Tinas Calories Burned,

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Step-by-step explanation:

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Step-by-step explanation:

[tex]

\underline{\bf{Given\::}}

Given:

\underline{\bf{To\:find\::}}

Tofind:

\underline{\bf{Explanation\::}}

Explanation:

\boxed{\bf{\frac{1}{f} =\frac{1}{v} -\frac{1}{u} }}}}

\begin{gathered}\longrightarrow\sf{\dfrac{1}{-10} =\dfrac{1}{v} -\dfrac{1}{-30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{1}{-10} +\dfrac{1}{30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{-3+1}{30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\cancel{\dfrac{-2}{30} }}\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{1}{-15} }\\\\\\\longrightarrow\sf{v=-15\:cm}\end{gathered}

−10

1

=

v

1

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1

v

1

=

−10

1

+

30

1

v

1

=

30

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v

1

=

30

−2

v

1

=

−15

1

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\boxed{\bf{M \:A \:G \:N\: I \:F \:I \:C\: A\: T \:I \:O\: N :}}

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↦m=

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Heightofimage(I)

=

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=

u

v

↦m=

−15

−30

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[/tex]

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Step-by-step explanation:

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