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2 years ago

What is -6/5 divided by 1/2

Mathematics

1 answer:

Elden [556K]2 years ago

8 0

Answer:

- 3

Step-by-step explanation:

You might be interested in

If John drove 210 miles in 3 hours, how far will he have driven in 4.5 hours?

VashaNatasha [74]

Answer:

3hours and 15 mins

Step-by-step explanation:

210 divied by 3=70

70=per hour

4 hours=280

70 divied by 2= 35

.5=35

280+35=315

315 miles in 4.5 hours

3 0

3 years ago

3 3/5 + 9 3/5 answer with a mixed number in simplest form

Romashka [77]

Both fractions have the same denominator so it's going to be easy. First set up your equation:

$3 \frac{3}{5} + 9 \frac{3}{5}$

Add the whole numbers:

$3 + 9 = 12$

Now add the fractions:

$\frac{3}{5} + \frac{3}{5} = \frac{6}{5}$

add both together:

$12 + \frac{6}{5} = 12 \frac{6}{5}$

$\frac{6}{5}$ is an improper fraction so change it:

$12 \frac{6}{5} = 13 \frac{1}{5}$

Since 6 is one more than 5, add 1 to the whole number and subtract the numerator and denominator(6 - 5 = 1) and make the remaining the new numerator. That leaves you with $13 \frac{1}{5}$

Your answer is $13 \frac{1}{5}$

3 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

Solving Graphically... PLEASE HELP

olasank [31]

Infinitely many? Arrows indicate that the graph goes on continuously without a stopping point, even in parabolas.

3 0

2 years ago

15x + 45 = 15 (3 + x) always never one solution how do I do it

const2013 [10]

Answer:

Step-by-step explanation:

First, you need to try to get rid of the ().

To do this:

15 * 3 and 15 * x

This makes 45 and 15x after the = , so the formula is now:

15x + 45 = 45 + 15x

Now get the 15x to the left side. When a number switches sides, it becomes negative when positive and positive when negative. So in this case 15x will become -15x and the 45(on the left) becomes - 45

15x - 15x = 45 - 45

Answer: 0

8 0

3 years ago

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