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2 years ago

A film begins at 21 25 and lasts for 100 minutes. At what time will it end? Express your answer using the 24- hour clock.

Mathematics

2 answers:

Softa [21]2 years ago

8 0

Answer:

23 05

Step-by-step explanation:

change 100 minutes to hours ,then add it to the starting time for the film.

Mars2501 [29]2 years ago

3 0

Answer:

23 05

Step-by-step explanation:

from 21 25-22 00 35min are passed

and remain With 65min from 100min

from 22 00-23 00 60min are passed

and remain With 5min so then at 23 05 all 100min are passed from 21 25. thanks

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Read 2 more answers

How can i differentiate this equation?

Dmitry_Shevchenko [17]

$\bf y=\cfrac{2x^2-10x}{\sqrt{x}}\implies y=\cfrac{2x^2-10x}{x^{\frac{1}{2}}} \\\\\\ \cfrac{dy}{dx}=\stackrel{\textit{quotient rule}}{\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2}x^{-\frac{1}{2}} \right)}{\left( x^{\frac{1}{2}} \right)^2}} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2\sqrt{x}} \right)}{\left( x^{\frac{1}{2}} \right)^2} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x}$

$\bf\cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{ \frac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2\sqrt{x}}}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2x\sqrt{x}}$

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