All categories

3 years ago

I have a big question what is 720 x 0.1

Mathematics

2 answers:

Basile [38]3 years ago

8 0

Answer:

Step-by-step explanation:

I hope this helps :)

zhannawk [14.2K]3 years ago

6 0

Answer:

Step-by-step explanation:

You might be interested in

nick has$7.00.Bagels cost $0.75 each,and a small container of cream cheese costs $1.29. write an iquality to find the number of

Arte-miy333 [17]

Let Nick can buy x number of bangels and y number of

6 0

1 year ago

to design the interior surface of a huge stainless-steel tank, you revolve the curve y=x^2 on the interval 0x4 about the y-axis.

Oksana_A [137]

Answer:

W = 21.44*10⁶J

Step-by-step explanation:

Given

y = x² (0 < x < 4)

γ = 10000 N/m³

W = ?

If we apply

y = 4² = 16 = h

y = x² ⇒ x = √y

then

V = π*(√y)²*dy = π*y*dy

W = (γ*V) *(16-y) = γ*π*y*dy*(16-y)

⇒ W = γ*π*∫y*(16-y)dy = γ*π*(8y²-(y³/3))

finally we obtain (0 < y < 16)

W = γ*π*(8y²-(y³/3)) = 10000*π*(8*16²-(16³/3)) = 21.44*10⁶J

7 0

3 years ago

10. A car repair company charges a $15 for an

Valentin [98]

$75

If the fixed cost is $15, it can be removed from the two hour charge to give the charge of ONLY two hours, which is $150.

This means the hourly charge is $150÷2, which is $75

8 0

3 years ago

Plz, HELP me with this question! Plz, HELP me with this question!

defon

C is the correct answer

4 0

3 years ago

Verify that the function u(x, y, z) = log x^2 + y^2 is a solution of the two dimensional Laplace equation u_xx + u_yy = 0 everyw

Daniel [21]

Answer:

The function $u(x,y,z)=log ( x^{2} +y^{2})$ is indeed a solution of the two dimensional Laplace equation $u_{xx} +u_{yy} =0$ .

The wave equation $u_{tt} =u_{xx}$ is satisfied by the function $u(x,t)=cos(4x)cos(4t)$ but not by the function $u(x,t)=f(x-t)+f(x+1)$ .

Step-by-step explanation:

To verify that the function $u(x,y,z)=log ( x^{2} +y^{2})$ is a solution of the 2D Laplace equation we calculate the second partial derivative with respect to x and then with respect to t.

$u_{xx}=\frac{2}{ln(10)}((x^{2} +y^{2})^{-1} -2x^{2} (x^{2} +y^{2})^{-2})$

$u_{yy}=\frac{2}{ln(10)}((x^{2} +y^{2})^{-1} -2y^{2} (x^{2} +y^{2})^{-2})$

then we introduce it in the equation $u_{xx} +u_{yy} =0$

we get that $\frac{2}{ln(10)} (\frac{2}{(x^{2}+y^{2}) } - \frac{2}{(x^{2}+y^{2} ) } )=0$

To see if the functions 1) $u(x,t)=cos(4x)cos(4t)$ and 2) $u(x,t)=f(x-t)+f(x+1)$ solve the wave equation we have to calculate the second partial derivative with respect to x and the with respect to t for each function. Then we see if they are equal.

1) $u_{xx}=-16 cos (4x) cos (4t)$

$u_{tt}=-16cos(4x)cos(4t)$

we see for the above expressions that $u_{tt} =u_{xx}$

2) with this function we will have to use the chain rule

If we call $s=x-t$ and $w=x+1$ then we have that

$u(x,t)=f(x-t)+f(x+1)=f(s)+f(w)$

So $\frac{\partial u}{\partial x}=\frac{df}{ds}\frac{\partial s}{\partial x} +\frac{df}{dw} \frac{\partial w}{\partial x}$

because we have $\frac{\partial s}{\partial x} =1$ and $\frac{\partial w}{\partial x} =1$

then $\frac{\partial u}{\partial x} =f'(s)+f'(w)$

⇒ $\frac{\partial^{2} u }{\partial x^{2} } =\frac{\partial}{\partial x} (f'(s))+ \frac{\partial}{\partial x} (f'(w))$

⇒ $\frac{\partial^{2} u }{ \partial x^{2} } =\frac{d}{ds} (f'(s))\frac{\partial s}{\partial x} +\frac{d}{ds} (f'(w))\frac{\partial w}{\partial x}$

⇒ $\frac{\partial^{2} u }{ \partial x^{2} } =f''(s)+f''(w)$

Regarding the derivatives with respect to time

$\frac{\partial u}{\partial t}=\frac{df}{ds} \frac{\partial s}{\partial t}+\frac{df}{dw} \frac{\partial w}{\partial t}=-\frac{df}{ds} =-f'(s)$

then $\frac{\partial^{2} u }{\partial t^{2} } =\frac{\partial}{\partial t} (-f'(s))=-\frac{d}{ds} (f'(s))\frac{\partial s}{\partial t} =f''(s)$

we see that $\frac{\partial^{2} u }{ \partial x^{2} } =f''(s)+f''(w) \neq f''(s)=\frac{\partial^{2} u }{\partial t^{2} }$

$u(x,t)=f(x-t)+f(x+1)$ doesn´t satisfy the wave equation.

4 0

3 years ago