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swat32
3 years ago
9

Can anyone help me solve this?​

Mathematics
2 answers:
erma4kov [3.2K]3 years ago
8 0

\\ \rm{:}\rightarrowtail sin^245°-tan^260°+cos^290°

  • sin45=1/√2
  • tan60=√3
  • cos90=1

\\ \rm{:}\rightarrowtail (\dfrac{1}{\sqrt{2}})^2-(\sqrt{3})^2+(0)^2

\\ \rm{:}\rightarrowtail \dfrac{1}{2}-3+0

\\ \rm{:}\rightarrowtail 1/2-3

\\ \rm{:}\rightarrowtail -5/2

Likurg_2 [28]3 years ago
3 0

Answer:

  • - 2.5

Step-by-step explanation:

  • sin²45° - tan²60° + cos²90° =
  • (√2/2)² - (√3)² + 0² =
  • 1/2 - 3 + 0 =
  • 0.5 - 3 =
  • - 2.5
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The rate at which the temperature is dropping is 6t 5t2 degrees Celsius per hour t hours after sundown. How much had the tempera
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Using a differential equation, it is found that the temperature dropped 72 degrees Celsius between sundown and 3 hours after sundown.

<h3>What is the differential equation that describes the temperature in t hours after sundown?</h3>

The rate at which the temperature is dropping is 6t + 5t^2 degrees Celsius per hour t hours after sundown, hence the <em>differential equation</em> is:

\frac{dT}{dt} = 6t + 5t^2

Applying <em>separation of variables</em>, we find the solution as follows:

\frac{dT}{dt} = 6t + 5t^2

dT = (6t + 5t^2) dt

\int dT = \int (6t + 5t^2) dt

T(t) = \frac{5t^3}{3} + 3t^2 + T(0)

In which T(0) is the temperature at sundown.

In 3 hours, the change will be of:

C = T(3) - T(0) = \frac{5t^3}{3} + 3t^2|_{t = 3}

Hence:

\frac{5(3)^3}{3} + 3(3)^2 = 45 + 27 = 72

The temperature dropped 72 degrees Celsius between sundown and 3 hours after sundown.

You can learn more about differential equations at brainly.com/question/24348029

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