<img src="https://tex.z-dn.net/?f=32.5%20%3D%2030%20%2B%20%20%5Cfrac%7B20%20-%20%2840%20%2B%20x%29%7D%7B12%7D%20%20%5Ctimes%2010

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3 years ago

" id="TexFormula1" title="32.5 = 30 + \frac{20 - (40 + x)}{12} \times 10" alt="32.5 = 30 + \frac{20 - (40 + x)}{12} \times 10" align="absmiddle" class="latex-formula">
please i need a step by step solution...I'll mark brainliest

Mathematics

1 answer:

erastova [34]3 years ago

7 0

Answer:

$x = - 23$

Step-by-step explanation:

$\frac{20 - (40 + x)}{12} \times 10 = 32.5 - 30 = 2.5$

$= > \frac{20 - (40 + x)}{12} = 0.25$

$= > \frac{20 - (40 + x)}{12} = \frac{1}{4}$

$= > 20 - (40 + x) = \frac{12}{4} = 3$

$= > 40 + x = 20 - 3 = 17$

$= > x = 17 - 40 = - 23$

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4 years ago

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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.

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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 $\frac{\partial w}{\partial x} =1$

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

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

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.

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3 years ago