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Crazy boy [7]
3 years ago
10

A trip 50 miles out of town takes 45 minutes. If the same person

Mathematics
1 answer:
dsp733 years ago
6 0

Hey there! I'm happy to help!

We see that it takes 45 minutes for a person to drive 50 miles. We can write this as a fraction that is 45/50, which simplifies to 9/10, meaning it would take this person 9 minutes to travel 10 miles.

So, how long would it take to travel 120? Well, we know that if we take 10 miles and multiply it by 12 we will have 120 miles. If we take the time it takes to drive those ten miles (9 minutes) and multiply it by 12, we will figure out how long it takes to drive 120 miles!

9×12=108

However, we want this to be written in hours. We know that there are 60 minutes in an hour, and if we subtract 60 from 108 we have 48. This gives us 1 hour and 48 minutes.

Therefore, it will take 1 hour and 48 minutes for this person to travel 120 miles at the same rate.

Have a wonderful day! :D

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Answer:

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Step-by-step explanation:

Here, the given function is p(x)=3x^5+2x^2 - 5

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Now, if ( 3 x +  5) = 0,

we get x = - 5/3

So, the zero of the given polynomial is x = -5/3

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Now, let us find the value of function at x = -5/3

Substitute x = -5/3 in the given function p(x), we get:

p(x)=3x^5+2x^2 - 5  \implies p(\frac{-5}{3})  = 3(\frac{-5}{3})^5 + 2(\frac{-5}{3})^2 - 5\\= 3(\frac{-3,125}{243}) + 2(\frac{25}{9})  - 5\\= (\frac{-3,125}{81}) + (\frac{50}{9})  - 5\\= -38.580 + 5.56  - 5  =  -38.02\\\implies p(\frac{-5}{3})  = -38.02

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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

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b = -1/37

Use this to find 'a'

a = 6b

a = 6(-1/37)

a = -6/37

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