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sdas [7]
2 years ago
9

Find the equation of tangent line

Mathematics
1 answer:
bogdanovich [222]2 years ago
3 0

\frac{dy}{dx} 2x {e}^{x}  = 2x {e }^{x}  + 2 {e}^{x}

Evaluated at x=0 we get

y = 2x

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How to find the area
sp2606 [1]
You know from trigonometry that AB = 15 units, so the area of the larger triangle ACB is (1/2)*15*15 = 112.5 units^2.

Then the area of triangle ADE is
.. (112.5 -62.5) = 50 . . . . units^2
7 0
3 years ago
A boy flying a kite lets out 300 feet of string, which makes an angle of 38o with the ground. Assuming that the string is straig
Greeley [361]

Answer:

The answer is 184.7 ft

Step-by-step explanation:

You need to find the distance from kite to ground

sin38°/1 = x/300

x = 300 · sin 38

Therefore your answer is 184.7

This is the best way I could explain it....Hope this helps:)

4 0
2 years ago
Quick, anyone know this geometry question?
scoray [572]
RS is 68 and T is 92 if you want an explanation then i gotten but u said quick so maybe it is timed idk
4 0
3 years ago
Read 2 more answers
4. 4c - 3 (c = -2)<br><br><br> will give 5 stars
Flauer [41]
4(-2)-3 = -8-3 = -11
Answer: -11
4 0
2 years ago
CALCULUS: For an object whose velocity in ft/sec is given by v(t) = sin(t), what is its distance, in feet, travelled on the inte
rodikova [14]

The linked answer is wrong because that integral gives you the net displacement of the object, not the total distance.

To get the distance, you have to integrate the speed (as opposed to velocity), which involves integrating the absolute value of the velocity function.

\mathrm{distance} = \displaystyle\int_1^5 |\sin(t)| \,\mathrm dt

By definition of absolute value,

|\sin(t)|=\begin{cases}\sin(t)&\text{for }\sin(t)\ge0\\-\sin(t)&\text{for }\sin(t)

Over this particular integration interval,

• sin(<em>t</em> ) ≥ 0 for 1 ≤ <em>t</em> < <em>π</em>, and

• sin(<em>t</em> ) < 0 for <em>π</em> < <em>t</em> ≤ 5

so you end up splitting the integral at <em>t</em> = <em>π</em> as

\mathrm{distance} = \displaystyle\int_1^\pi \sin(t)\,\mathrm dt + \int_\pi^5 (-\sin(t))\,\mathrm dt

Now compute the distance:

\mathrm{distance} = -\cos(t)\bigg|_1^\pi + \cos(t)\bigg|_\pi^5

\mathrm{distance} = -(\cos(\pi) - \cos(1)) + (\cos(5) - \cos(\pi))

\mathrm{distance} = -2\cos(\pi) + \cos(1) + \cos(5) \approx 2.82

making B the correct answer.

7 0
2 years ago
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