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2 years ago

I will mark brainiest:

Mathematics

1 answer:

Alla [95]2 years ago

7 0

Answer:

negative

Step-by-step explanation:

because a positive multiplied by a negative number equals a negative number.

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Dvinal [7]

B cause you take y times it by the circumference

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2 years ago

Which of the following could not be points on the unit circle?

NemiM [27]

A. √(0.8^2) + (0.6^2) = √1 = 1 => OK

<span>b.(-2/3,√ 5/3) = √(-2/3)^2 + 5/9) = √(4/9 +5/9) = √1 = 1 => OK

c.(√ 3/2, 1/3) = √(3/4 + 1/9) < 1 => it is inside the unit circle

d.(1,1) = √(1 + 1) = √2 > 1 => NO. This point is beyond the limits of the unit circle.</span>

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3 years ago

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Please help me find the area of this irregular shape.

ANTONII [103]

Answer:

23 Cm(2)

Step-by-step explanation:

Sincerest Apologies if it is incorrect

if it's not 23 it may be:

26 Cm(2)

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2 years ago

Anybody? I'm pretty sure calculators don't do it right so..please no calculators. ;-;

Dima020 [189]

Answer:

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2 years ago

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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