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Paha777 [63]
3 years ago
7

please help. No links or fake answers. If your going to answer, may you please do one screenshot at the least. Thank you for you

r time.

Mathematics
1 answer:
Lady bird [3.3K]3 years ago
7 0

Answer:

29

Step-by-step explanation:

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The amount a spring will stretch, S, varies directly with the weight, F, attached to the spring. If
nexus9112 [7]

Answer: 1.5 inches

Step-by-step explanation:

If two quantities are directly proportional, it means that an increase in the value of one quantity would cause a corresponding increase in the other quantity. Also, a decrease in the value of one quantity would cause a corresponding decrease in the other quantity.

Given that the amount a spring will stretch,S varies directly with weight, F, if we introduce a constant of proportionality, k, the expression becomes

S = kF

When S = 2.22 and L = 37, then

2.22 = 37k

k = 2.22/37 = 0.06

Therefore, the equation that relates the length S that the spring stretches to the weight F is

S = 0.06F

Therefore, if F = 25 pounds, then

S = 0.06 × 25

S = 1.5 inches

6 0
3 years ago
A new school will have three playgrounds. Two playgrounds will be large and one playground will be small. The district administr
Andrews [41]

Answer:

you multiple two numbers and you devide both sides you get the answer

7 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} )

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