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2 years ago

Plzz answer fast I know what it is but I need to double check!!!!!

Mathematics

2 answers:

alukav5142 [94]2 years ago

4 0

Answer:

it's B

Step-by-step explanation:

just solve them and they match

irina1246 [14]2 years ago

3 0

Answer:

If your thinking B is the answer then you are right

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It takes a train 60 seconds to completely drive through a bridge of 1260 m, and 90 seconds to completely drive through a tunnel

Olegator [25]

The length of train is 120 meters

The speed of train is 25 meter per second

Solution:

The speed is given by formula:

$speed = \frac{distance}{time}$

Given that,

It takes a train 60 seconds to completely drive through a bridge of 1260 m

And 90 seconds to completely drive through a tunnel of 2010 m

Let the length of the train be "x"

The total distance travelled by the train across the bridge is given by:

total distance = x + 1260 + x = 2x + 1260

[ here we use x + x to represent that train completely drives through bridge ]

The time taken is given as 60 seconds

Therefore, the speed is given as:

$speed = \frac{2x+ 1260}{60}$

The total distance travelled by the train through the tunnel is given by:

total distance = x + 2010 + x = 2x + 2010

The time taken is given as 90 seconds

Therefore, the speed is given as:

$speed = \frac{2x+ 2010}{90}$

The speed of the train was the same in both cases.

Equate both speeds

$\frac{2x+ 1260}{60} = \frac{2x+ 2010}{90}$

$\frac{x+630}{30} = \frac{x+1005}{45}$

$45(x+630) = 30(x+1005)\\\\45x + 28350 = 30x + 30150\\\\45x - 30x = 30150 - 28350\\\\15x = 1800\\\\x = 120$

Thus length of train is 120 meters

Find the speed of train:

$speed = \frac{2x+ 1260}{60}\\\\speed = \frac{2(120) + 1260)}{60}\\\\speed = 25$

Thus speed of train is 25 meter per second

5 0

2 years ago

In simplified exponential notation, the expression a²·a -3·a= A 1/a B 0 C 1

dalvyx [7]

Answer a^3-3a

a^2 times a = a^3 you add exponents

4 0

3 years ago

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Does anyone know all the concepts/rules needed to learn math in college? I want to make sure I have things I need now in my seni

Mandarinka [93]

Answer:

Basic college math topics include whole numbers, fractions, numbers, decimals and integers. Problem solving, algebra, percents and geometry are also included in course content.

Step-by-step explanation:

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2 years ago

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an airplane has flown 260 miles out of a total trip of 500 miles. What fraction in simplest form of the trip has been completed

solmaris [256]

260/500 = .52
.52 = 13/25
.52 = 52%

4 0

2 years ago

Solve the following differential equation using using characteristic equation using Laplace Transform i. ii y" +y sin 2t, y(0) 2

kifflom [539]

Answer:

The solution of the differential equation is $y(t)= - \frac{1}{3} Sin(2t)+2 Cos(t)+\frac{5}{3} Sin(t)$

Step-by-step explanation:

The differential equation is given by: y" + y = Sin(2t)

i) Using characteristic equation:

The characteristic equation method assumes that y(t)= $e^{rt}$ , where "r" is a constant.

We find the solution of the homogeneus differential equation:

y" + y = 0

$y'=re^{rt}$

$y"=r^{2}e^{rt}$

$r^{2}e^{rt}+e^{rt}=0$

$(r^{2}+1)e^{rt}=0$

As $e^{rt}$ could never be zero, the term (r²+1) must be zero:

(r²+1)=0

r=±i

The solution of the homogeneus differential equation is:

$y(t)_{h}=c_{1}e^{it}+c_{2}e^{-it}$

Using Euler's formula:

$y(t)_{h}=c_{1}[Sin(t)+iCos(t)]+c_{2}[Sin(t)-iCos(t)]$

$y(t)_{h}=(c_{1}+c_{2})Sin(t)+(c_{1}-c_{2})iCos(t)$

$y(t)_{h}=C_{1}Sin(t)+C_{2}Cos(t)$

The particular solution of the differential equation is given by:

$y(t)_{p}=ASin(2t)+BCos(2t)$

$y'(t)_{p}=2ACos(2t)-2BSin(2t)$

$y''(t)_{p}=-4ASin(2t)-4BCos(2t)$

So we use these derivatives in the differential equation:

$-4ASin(2t)-4BCos(2t)+ASin(2t)+BCos(2t)=Sin(2t)$

$-3ASin(2t)-3BCos(2t)=Sin(2t)$

As there is not a term for Cos(2t), B is equal to 0.

So the value A=-1/3

The solution is the sum of the particular function and the homogeneous function:

$y(t)= - \frac{1}{3} Sin(2t) + C_{1} Sin(t) + C_{2} Cos(t)$

Using the initial conditions we can check that C1=5/3 and C2=2

ii) Using Laplace Transform:

To solve the differential equation we use the Laplace transformation in both members:

ℒ[y" + y]=ℒ[Sin(2t)]

ℒ[y"]+ℒ[y]=ℒ[Sin(2t)]

By using the Table of Laplace Transform we get:

ℒ[y"]=s²·ℒ[y]-s·y(0)-y'(0)=s²·Y(s) -2s-1

ℒ[y]=Y(s)

ℒ[Sin(2t)]= $\frac{2}{(s^{2}+4)}$

We replace the previous data in the equation:

s²·Y(s) -2s-1+Y(s) = $\frac{2}{(s^{2}+4)}$

(s²+1)·Y(s)-2s-1= $\frac{2}{(s^{2}+4)}$

(s²+1)·Y(s)= $\frac{2}{(s^{2}+4)}+2s+1=\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)}$

Y(s)= $\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)(s^{2}+1)}$

Y(s)= $\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}$

Using partial franction method:

$\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}=\frac{As+B}{s^{2}+4} +\frac{Cs+D}{s^{2}+1}$

2s^{3}+s^{2}+8s+6=(As+B)(s²+1)+(Cs+D)(s²+4)

2s^{3}+s^{2}+8s+6=s³(A+C)+s²(B+D)+s(A+4C)+(B+4D)

We solve the equation system:

A+C=2

B+D=1

A+4C=8

B+4D=6

The solutions are:

A=0 ; B= -2/3 ; C=2 ; D=5/3

So,

Y(s)= $\frac{-\frac{2}{3} }{s^{2}+4} +\frac{2s+\frac{5}{3} }{s^{2}+1}$

Y(s)= $-\frac{1}{3} \frac{2}{s^{2}+4} +2\frac{s }{s^{2}+1}+\frac{5}{3}\frac{1}{s^{2}+1}$

By using the inverse of the Laplace transform:

ℒ⁻¹[Y(s)]=ℒ⁻¹[ $-\frac{1}{3} \frac{2}{s^{2}+4}$ ]-ℒ⁻¹[ $2\frac{s }{s^{2}+1}$ ]+ℒ⁻¹[ $\frac{5}{3}\frac{1}{s^{2}+1}$ ]

$y(t)= - \frac{1}{3} Sin(2t)+2 Cos(t)+\frac{5}{3} Sin(t)$

3 0

3 years ago