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3 years ago

Find the length of AB. Leave

Mathematics

1 answer:

vovikov84 [41]3 years ago

3 0

Answer:

AB = 3π

Step-by-step explanation:

See attachment for correct format of question.

Given

From the attachment, we have that

θ = 20°

Radius, r = 27

Required

Find length of AB

AB is an arc and it's length can be calculated using arc length formula.

$Length = \frac{\theta}{360} * 2\pi r$

Substitute 20 for θ and 27 for r

$Length = \frac{20}{360} * 2\pi *27$

$Length = \frac{20 * 2\pi * 27}{360}$

$Length = \frac{1080 \pi}{360}$

$Length = 3\pi$

Hence, the length of arc AB is terms of π is 3π

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Step-by-step explanation:

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2 years ago

Determine whether the two expressions are equivalent. Explain your reasoning.

dlinn [17]

Let us evaluate each pair of expressions one at a time.
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2(x + 3) = 2x + 6 which is not equal to 3x + 5.
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10. 15 - 6x and 15(1 - 6x)
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Answer: Only the expressions in number 11 are equivalent.

6 0

2 years ago

Simplify -3p 3 + 5p + (-2p 2) + (-4) - 12p + 5 - (-8p 3).

MissTica

Answer:5p^3-2p^2-7p+1

Step-by-step explanation:

-3p^3+5p+(-2p^2)+(-4)-12p+5-(-8p^3)

open brackets

-3p^3+5p-2p^2-4-12p+5+8p^3

Collect like terms

-3p^3+8p^3-2p^2+5p-12p-4+5

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3 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

What is the Laplace Transform of 7t^3 using the definition (and not the shortcut method)

Leokris [45]

Answer:

Step-by-step explanation:

By definition of Laplace transform we have

L{f(t)} = $L{{f(t)}}=\int_{0}^{\infty }e^{-st}f(t)dt\\\\Given\\f(t)=7t^{3}\\\\\therefore L[7t^{3}]=\int_{0}^{\infty }e^{-st}7t^{3}dt\\\\$

Now to solve the integral on the right hand side we shall use Integration by parts Taking $7t^{3}$ as first function thus we have

$\int_{0}^{\infty }e^{-st}7t^{3}dt=7\int_{0}^{\infty }e^{-st}t^{3}dt\\\\= [t^3\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(3t^2)\int e^{-st}dt]dt\\\\=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\$

Again repeating the same procedure we get

$=0-\int_{0}^{\infty }\frac{3t^{2}}{-s}e^{-st}dt\\\\=\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt\\\\\int_{0}^{\infty }\frac{3t^{2}}{s}e^{-st}dt= \frac{3}{s}[t^2\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t^2)\int e^{-st}dt]dt\\\\=\frac{3}{s}[0-\int_{0}^{\infty }\frac{2t^{1}}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{2}}[\int_{0}^{\infty }te^{-st}dt]\\\\$

Again repeating the same procedure we get

$\frac{3\times 2}{s^2}[\int_{0}^{\infty }te^{-st}dt]= \frac{3\times 2}{s^{2}}[t\int e^{-st} ]_{0}^{\infty}-\int_{0}^{\infty }[(t)\int e^{-st}dt]dt\\\\=\frac{3\times 2}{s^2}[0-\int_{0}^{\infty }\frac{1}{-s}e^{-st}dt]\\\\=\frac{3\times 2}{s^{3}}[\int_{0}^{\infty }e^{-st}dt]\\\\$

Now solving this integral we have

$\int_{0}^{\infty }e^{-st}dt=\frac{1}{-s}[\frac{1}{e^\infty }-\frac{1}{1}]\\\\\int_{0}^{\infty }e^{-st}dt=\frac{1}{s}$

Thus we have

$L[7t^{3}]=\frac{7\times 3\times 2}{s^4}$

where s is any complex parameter

5 0

2 years ago