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Leto [7]
3 years ago
14

Does anyone know this lol i just need a short explanation

Mathematics
1 answer:
maks197457 [2]3 years ago
4 0

Answer: pretty sure it’s because the left and bottom sides add up to equal the top side

Step-by-step explanation:

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A bucket contains 11 2/3 gallons of water and 5/6 is full. How many gallons of water would be in a full bucket? Write a multipli
vagabundo [1.1K]

Answer:

14 gallons

Step-by-step explanation:

If 11 2/3 gallons represents 5/6 full bucket then we can equate that

x=1 full

x=\frac {11 2/3}{5/6}=14

Therefore, the bucket requires 14 gallons

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3 years ago
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VerZion sells the iPhone X for $899. If the monthly plan for service is $65 per month what is the linear equation for the cost o
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899$ is the y intercept because that is your starting value. Then, 65 is your slope. (Rise 65, run 1). After 24 months, the total cost will be 2459$. This is because 24 • 65 = 1560. This is your total cost of phone service. Then, you need to add the phone. 1560 + 899 = 2459. So you total cost is 2459.
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What is central and inscribed angle​
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Answer:

An inscribed angle is an angle formed by two chords in a circle which have a common endpoint. This common endpoint forms the vertex of the inscribed angle. The other two endpoints define what we call an intercepted arc on the circle. The intercepted arc might be thought of as the part of the circle which is "inside" the inscribed angle.

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3 years ago
Given: abcd is a rectangle. prove: abcd has congruent diagonals. identify the steps that complete the proof.
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Parallel side theorem 
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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
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