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77julia77 [94]
3 years ago
12

Find the distance between these points. C(0, 4), T(-6, -3) √(37) √(85) √(109)

Mathematics
2 answers:
Leona [35]3 years ago
5 0
<span>√85</span>≈<span>9.21954445

Hope this helps. c:</span>
lorasvet [3.4K]3 years ago
5 0

Answer:

(B) CT=\sqrt{85}

Step-by-step explanation:

Given: Two points C(0,4) and T(-6,-3).

To find: The distance between the given points C(0,4) and T(-6,-3).

Solution: It is given taht the two points are C(0,4) and T(-6,-3), we have to find the distance between these two points, thus using teh distance formula, we have:

CT=\sqrt{(-6-0)^2+(-3-4)^2}

⇒CT=\sqrt{36+49}

⇒CT=\sqrt{85}

which is the required distance between the two given points.

Therefore, option B is correct.

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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
Whitepunk [10]

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

5 0
2 years ago
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