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Allisa [31]
3 years ago
12

Lily wants to buy a smartphone, she has $500 to spend .list three ways she can get a good deal

Mathematics
1 answer:
BaLLatris [955]3 years ago
3 0
1. Check the prices of phones from different stores.

2. Do some research to see what is the average cost of the phone.

3. Check for possible discounts.

Hope this helps :)

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Use the Laplace transform to solve the given initial value problem. y' + 6y = e^4t ; y(0)=2 ...?
TiliK225 [7]
Performing laplace transform of the equation.

sY(s) - y(0) + 6Y(s) = 1/(s-4)
(s+6)Y(s) - 2 = 1/(s-4)
Y(s) = 2/(s+6) + 1/(s-4)(s+6), by partial fraction decomposition
Y(s) = 2/(s+6) + 1/10 * (1/(s-4) + 1/(s+6))
Y(s) = 0.1/(s-4) + 2.1/(s+6)

Performing inverse laplace transform,
y(t) = 0.1e^4t + 2.1e^(-6t)


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
4 0
3 years ago
Two fractions negative whose product is 5/8
Andreyy89
You could do -1/-2 times -5/-4 which would be 5/8
8 0
3 years ago
A city has a population of 250,000 people. Suppose that each year the population grows by 7.5%. What will the population be afte
Diano4ka-milaya [45]
So, if it grows by 6.75%, each year the population is 106.75% of the year before.

After 1 year, 370,000(1.0675). After two years, 370,000(1.0675)(1.0675).

370,000(1.0675)12 = your answer
4 0
3 years ago
The chart below shows conversion between kilometers and miles.
Feliz [49]

Answer:

12

Step-by-step explanation:

1.2 divided by 2 is 0.6 and you want to know how much twenty is bc thats the whole point of the question lol

but anyway multiply 0.6 and 20 to get 12

yw

6 0
4 years ago
Read 2 more answers
At noon, ship A is 130 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 20 km/h. How fast is
Hitman42 [59]

Answer:

answer = 12.87 km/h

Step-by-step explanation:

Given

Ship A is sailing east at 25 km/h = \frac{dx}{dt}

ship B is sailing north at 20 km/h =\frac{dy}{dt}

here x and y are the  sailing at t = 4 : 00 pm for ship A and B respectively

so we get x = 4 ×25 =100 km/h

                 y = 4× 20 = 80 km/h

let z is the distance between the ships, we need to find \frac{dz}{dt} at t = 4 hr

At noon, ship A is 130 km west of ship B (12:00 pm)

so equation will be

z^2 = (130-x)^2 + y^2......................(i)\\put x = 100 and y = 80 \\\\we |  | get \\

z^2 = 30^2 + 80^2\\z =\sqrt{7300} km/h

derivative first equation w . r. to t we get

2z\frac{dz}{dt} =-2(130-x)\frac{dx}{dt}+2y\frac{dy}{dt}

\frac{dz}{dt} =\frac{1}{z}[(x -130)\frac{dx}{dt} +y\frac{dy}{dt}]

\frac{dz}{dt} = \frac{( -20\times25 + 80\times20)}{\sqrt{7300} }

     = \frac{1100}{85.44}\\  = 12.87km/h

8 0
3 years ago
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