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Brut [27]
4 years ago
7

Which of the follow is NOT a way to prove triangles similar?

Mathematics
1 answer:
frozen [14]4 years ago
3 0

Answer:

D

Step-by-step explanation: dosnt need to be all congruent see the question says "similar" not ecactly the same

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I have been trying to firgue this one out for a long time I haven't gotten it sorry.

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What are the two types of percent of change
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Percentage errors and percentage difference
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(a) A fire station is to be located along a road of length A, A<
Marrrta [24]

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Answer in picture attached

Step-by-step explanation:

6 0
3 years ago
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
HELP ASAP NOW!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Sedbober [7]

Answer:

x +2y = 36

2x+3y =64

Step-by-step explanation:

x = price of a T-shirt

y = price of a puck

Adam buys a T-shirt and two souvenir pucks for $36.00

x +2y = 36

Luke spends $64.00 on two T-shirts and three pucks

64 = 2x+3y

Our system of equations is

x +2y = 36

2x+3y =64

6 0
3 years ago
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