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3 years ago

A (0,0) B(-6,0) C (-6, 3) (x,y) - (x + 6, y + 3)

Mathematics

1 answer:

juin [17]3 years ago

3 0

Answer:

plz send full question

Step-by-step explanation:

according to your book

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Furkat [3]

Answer:

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Step-by-step explanation:

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3 years ago

Find the exact length of the curve. x=et+e−t, y=5−2t, 0≤t≤2 For a curve given by parametric equations x=f(t) and y=g(t), arc len

Rama09 [41]

The length of a curve C parameterized by a vector function r(t) = x(t) i + y(t) j over an interval a ≤ t ≤ b is

$\displaystyle\int_C\mathrm ds = \int_a^b \sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2} \,\mathrm dt$

In this case, we have

x(t) = exp(t ) + exp(-t ) ==> dx/dt = exp(t ) - exp(-t )

y(t) = 5 - 2t ==> dy/dt = -2

and [a, b] = [0, 2]. The length of the curve is then

$\displaystyle\int_0^2 \sqrt{\left(e^t-e^{-t}\right)^2+(-2)^2} \,\mathrm dt = \int_0^2 \sqrt{e^{2t}-2+e^{-2t}+4}\,\mathrm dt$

$=\displaystyle\int_0^2 \sqrt{e^{2t}+2+e^{-2t}} \,\mathrm dt$

$=\displaystyle\int_0^2\sqrt{\left(e^t+e^{-t}\right)^2} \,\mathrm dt$

$=\displaystyle\int_0^2\left(e^t+e^{-t}\right)\,\mathrm dt$

$=\left(e^t-e^{-t}\right)\bigg|_0^2 = \left(e^2-e^{-2}\right)-\left(e^0-e^{-0}\right) = \boxed{e^2-\frac1{e^2}}$

5 0

3 years ago

Answer 1a please and thank you

arlik [135]

Answer:h

Step-by-step explanation:

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2 years ago

Plz help me with this

madreJ [45]

In order a0->a5: 1/3, 1, 3, 9, 27, 81

All you're doing is plugging the x-value in. So for a2, x=1, so y=3¹ which means y=3.

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3 years ago

Read 2 more answers

How do you find the slope and y-intercept

ankoles [38]

To find them it's y=mx+b

b=the y intercept
m=the slope

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4 years ago