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1 year ago

(Refer to the attached image)

Mathematics

1 answer:

DENIUS [597]1 year ago

5 0

<h3>Answer: 9 meters</h3>

===========================================

Work Shown:

side a = 8
side b = 5
angle C = 88 degrees

Applying the law of cosines

c^2 = a^2 + b^2 - 2*a*b*cos(C)

c^2 = 8^2 + 5^2 - 2*8*5*cos(88)

c^2 = 64 + 25 - 80*cos(88)

c^2 = 86.2080403

c = sqrt(86.2080403)

c = 9.2848285

c = 9

Technically he needs 10 meters of rope because of the extra 0.28 portion, but I'll stick with 9 meters since your teacher said to round to the nearest whole number.

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Consider the curve defined by the equation y=6x2+14x. Set up an integral that represents the length of curve from the point (−2,

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Answer:

32.66 units

Step-by-step explanation:

We are given that

$y=6x^2+14x$

Point A=(-2,-4) and point B=(1,20)

Differentiate w.r. t x

$\frac{dy}{dx}=12x+14$

We know that length of curve

$s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx$

We have a=-2 and b=1

Using the formula

Length of curve= $s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx$

Using substitution method

Substitute t=12x+14

Differentiate w.r t. x

$dt=12dx$

$dx=\frac{1}{12}dt$

Length of curve= $s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt$

We know that

$\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C$

By using the formula

Length of curve= $s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})$