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3 years ago

Hypotenuse of a right triangles whose legs are square root of 10 and sqare root of 15

2 answers:

6 0

Sqrt{10}² + sqrt15² = c²

The square of any square root is that number without the square root sign

So now we have 10 + 15= c²
25 = c²
So c= 5

wlad13 [49]3 years ago

3 0

10^2 + 15^2 = c^2 

100 + 225 = c^2 

325 = c^2 

Find the square root of both sides .... 

18.02 = c = hypotenuse.

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The cost of renting an apartment in the downtown city area is increasing by 4% each year. Currently, the average rent for

Novay_Z [31]

Answer:500

Step-by-step explanation:

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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

Use the formula for the cosine of the difference of two angles to find the exact value of the following expression.

Zolol [24]

Answer:

Exact value of Cos(45° - 60°) is 0.96 using difference of two angles.

Step-by-step explanation:

Given:

Cos(45° - 60°)

We have to apply the formula of cosine for difference of the two angles.

Formula:

$cos(a-b)=cos(a)\ cos(b)+sin(a)\ sin(b)$

Plugging the values.

⇒ $cos(45-60)=cos(45)\ cos(60) + sin(45)\ sin(60)$

We know that the values :

$sin(45) =cos(45) = \frac{1}{\sqrt{2} }$

$sin(60)=\frac{\sqrt{3} }{2}$ and $cos(60)=\frac{1}{2}$

So,

⇒ $cos(45-60)=(\frac{1}{\sqrt{2} } \times \frac{1}{2} ) + (\frac{1}{\sqrt{2} } \times \frac{\sqrt{3} }{2})$

⇒ $cos(45-60)=(\frac{1}{2\sqrt{2} } + \frac{\sqrt{3} }{2\sqrt{2} })$

⇒ $cos(45-60)=(\frac{1+\sqrt{3} }{2\sqrt{2} } )$

⇒ $cos(45-60)=(\frac{1+\sqrt{3} }{2\sqrt{2} } )\times \frac{2\sqrt{2} }{2\sqrt{2} }$ ...rationalizing

⇒ $cos(45-60)=\frac{2\sqrt{2} +2\sqrt{6} }{ 8}$

⇒ $cos(45-60)=\frac{2(\sqrt{2}+\sqrt{6})}{8}$ ...taking 2 as a common factor

⇒ $cos(45-60)=\frac{(\sqrt{2}+\sqrt{6})}{4}$

To find the exact values we have to put the values of sq-rt .

As, $\sqrt{2}=1.41$ and $\sqrt{6} =2.44$

Then

⇒ $cos(45-60)=\frac{( 1.41+2.44)}{4}$

⇒ $cos(45-60)=\frac{( 3.85)}{4}$

⇒ $cos(45-60)=0.96$

So the exact value of Cos(45° - 60°) is 0.96 using difference of two angles.

3 0

3 years ago

The force F of gravity on a rocket varies directly with its mass m and inversely with the square of its distance d from Earth. W

xz_007 [3.2K]

Model for the Combined variation is $F = \frac{km}{d^{2} }$

Equation to find the mass of the rocket in terms of F and d is :

$m=\frac{F d^{2} }{k}$

<h3>What is Force ?</h3>

A force is the quantity that change the motion of the object. Force is that influence which can cause any object having some mass, to change it's position or to accelerate from it's original position.

Force (F) and mass (m) of any object are directly proportional to each other. More the mass of the object , more the force will require to accelerate it.

Mathematically, we can write the equation of force as :

F = m . a

where , m = mass of the object

a = constant value

In the statement given,

As, force of gravity varies directly with its mass

Therefore, F= k m , where k is a constant

Also force is inversely varies with square of distance d

Thus, F = k / d²

Combining both equations :

$F = \frac{km}{d^{2} }$

and the equation of mass in terms of F and d can be written as :

$m=\frac{F d^{2} }{k}$

To know more about Force and it's equation, visit :

brainly.com/question/14857727

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10 months ago

Cecil had 6 ice cubes. He put 1 ice cube in each glass. In how many glasses did Cecil put ice cubes?

Kamila [148]

He only puts 1 in each glass and he has six. The answer is 6.

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3 years ago

Read 2 more answers