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Yuri [45]
2 years ago
15

The sum of -2 and three times a number

Mathematics
1 answer:
algol132 years ago
6 0
The sum (addition) of -2 and 3 times (3 ·) a number (n)

-2 + 3n
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Eurostar is a high-speed railway service connecting
erma4kov [3.2K]

Answer:

Step-by-step explanation:

This is simply a units conversion problem.  It gives us for the number of passengers, the number of seats per carriage and the number of carriages per train.  To change the units from passengers to trains without changing the value, we use the multiplicative identity (that is, 1).

350000 passengers

(350000 passengers) * 1

(350000 passengers) * ((1 carriage)/(32 passengers)) * ((1 train)/(15 carriages)

   [note:  passengers and carriages cancel. Leaving only trains]

(350000)*(1/32)*(1/15) trains       [note: I write this way to paste into MS Excel]

729.1667  trains                 [oh, but don’t just round this number either up or down]

   729 full trains can carry 729*32*15 = 349920 passengers

   730 full trains can carry 730*32*15 = 350400 passengers

Now, we can say that 730 trains are adequate to carry 350000 passengers.

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3 years ago
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Julio signs a new lease for $800 and in the agreement the complex will raise his rent 3% each year that he renews. As of right n
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2 years ago
Find constants a and b such that the function y = a sin(x) + b cos(x) satisfies the differential equation y'' + y' − 5y = sin(x)
vichka [17]

Answers:

a = -6/37

b = -1/37

============================================================

Explanation:

Let's start things off by computing the derivatives we'll need

y = a\sin(x) + b\cos(x)\\\\y' = a\cos(x) - b\sin(x)\\\\y'' = -a\sin(x) - b\cos(x)\\\\

Apply substitution to get

y'' + y' - 5y = \sin(x)\\\\\left(-a\sin(x) - b\cos(x)\right) + \left(a\cos(x) - b\sin(x)\right) - 5\left(a\sin(x) + b\cos(x)\right) = \sin(x)\\\\-a\sin(x) - b\cos(x) + a\cos(x) - b\sin(x) - 5a\sin(x) - 5b\cos(x) = \sin(x)\\\\\left(-a\sin(x) - b\sin(x) - 5a\sin(x)\right)  + \left(- b\cos(x) + a\cos(x) - 5b\cos(x)\right) = \sin(x)\\\\\left(-a - b - 5a\right)\sin(x)  + \left(- b + a - 5b\right)\cos(x) = \sin(x)\\\\\left(-6a - b\right)\sin(x)  + \left(a - 6b\right)\cos(x) = \sin(x)\\\\

I've factored things in such a way that we have something in the form Msin(x) + Ncos(x), where M and N are coefficients based on the constants a,b.

The right hand side is simply sin(x). So we want that cos(x) term to go away. To do so, we need the coefficient (a-6b) in front of that cosine to be zero

a-6b = 0

a = 6b

At the same time, we want the (-6a-b)sin(x) term to have its coefficient be 1. That way we simplify the left hand side to sin(x)

-6a  -b = 1

-6(6b) - b = 1 .... plug in a = 6b

-36b - b = 1

-37b = 1

b = -1/37

Use this to find 'a'

a = 6b

a = 6(-1/37)

a = -6/37

8 0
2 years ago
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