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Vadim26 [7]
4 years ago
5

R, S, and T are coplanar. True False

Mathematics
1 answer:
Pavel [41]4 years ago
8 0

False, because when you connect them the don't even look like a line.

Hope this helps! ;)

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Find the inverse laplace transform of: (2 s + 4) / (s - 3)^3
Serhud [2]

Answer:

e^{3t}(2t+5t^{2})

Step-by-step explanation:

L^{-1}[\frac{2s+4}{(s-3)^{3}} ]=

Using the Translation theorem to transform the s-3 to s, that means multiplying by and change s to s+3

Translation theorem:L^{1} [F(s-a)=L^{-1}[F(s)|_{s \to s-a}\\ L^{1} [F(s-a)=e^{at} f(t)

L^{-1}[\frac{2s+4}{(s-3)^{3}} ]=e^{3t} L^{-1}[\frac{2(s+3)+4}{s^{3}} ]

Separate the fraction in a sum:

e^{3t} L^{-1}[\frac{2s+10}{s^{3}} ]=e^{3t} L^{-1}[\frac{2s}{s^{3}}+\frac{10}{s^{3}} ]=e^{3t} (L^{-1}[\frac{2}{s^{2}}]+ L^{-1}[\frac{10}{s^{3}}])

The formula for this is:

L^{-1}[\frac{n!}{s^{n+1}} ]=t^{n}

Modify the expression to match the formula.

e^{3t} (2L^{-1}[\frac{1}{s^{1+1}}]+ \frac{10}{2} L^{-1}[\frac{2}{s^{2+1}}])=e^{3t} (2L^{-1}[\frac{1}{s^{1+1}}]+ 5 L^{-1}[\frac{2}{s^{2+1}}])

Solve

e^{3t} (2L^{-1}[\frac{1}{s^{1+1}}]+ 5 L^{-1}[\frac{2}{s^{2+1}}])=e^{3t}(2t+5t^{2} )

6 0
3 years ago
I really need help with these I have no brain when it comes to math T^T
umka21 [38]
Answer: 4th option

Explanation:

x = 2/3 • pi • r^3
x divide by 2/3 = pi • r^3
x • 3/2 = pi • r^3
3x/2 = pi • r^3
3x/2 divide by pi = r^3
3x/2 • 1/pi = r^3
3x/2(pi) = r^3
Cube root(3x/2(pi)) = r
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I PART B: Which quote from the text best supports the answer to Part
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Answer:

b

Step-by-step explanation:

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A rectangular garden has a perimeter of 100 feet. The length of the garden is 20 feet greater than twice the width. The system o
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The length is 40ft and the width is 10 ft
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Ayden had $125.35 saved up. He wanted to buy a surf board that costs 400. On his way to work he saw some snorkel gear and bought
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334.64 i think

Step-by-step explanation:

I DO MATHS

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