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kirza4 [7]
3 years ago
6

Terry sees this offer. 35% off refurbished phone, now only £78. What was the orginal price?

Mathematics
1 answer:
KengaRu [80]3 years ago
8 0
35% off means that now it costs 100%-35%=65% of the original price

<span>£78 ---- 65%
x------100%

x=78*100/65=120</span>£<span>

or 
part/whole=0.65
78/whole=0.65
whole =78/0.65 =120 </span>£ 
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Use the Chain Rule to find the indicated partial derivatives. u = x2 + yz, x = pr cos(θ), y = pr sin(θ), z = p + r; (partial u)/
devlian [24]

u(x,y,z)=x^2+yz

\begin{cases}x(p,r,\theta)=pr\cos\theta\\y(p,r,\theta)=pr\sin\theta\\z(p,r,\theta)=p+r\end{cases}

At the point (p,r,\theta)=(2,2,0), we have

\begin{cases}x(2,2,0)=4\\y(2,2,0)=0\\z(2,2,0)=4\end{cases}

Denote by f_x:=\dfrac{\partial f}{\partial x} the partial derivative of a function f with respect to the variable x. We have

\begin{cases}u_x=2x\\u_y=z\\u_z=y\end{cases}

The Jacobian is

\begin{bmatrix}x_p&x_r&x_\theta\\y_p&y_r&y_\theta\\z_p&z_r&z_\theta\end{bmatrix}=\begin{bmatrix}r\cos\theta&p\cos\theta&-pr\sin\theta\\r\sin\theta&p\sin\theta&pr\cos\theta\\1&1&0\end{bmatrix}

By the chain rule,

u_p=u_xx_p+u_yy_p+u_zz_p=2xr\cos\theta+zr\sin\theta+y

u_p(2,2,0)=2\cdot4\cdot2\cos0+4\cdot2\sin0+0\implies\boxed{u_p(2,2,0)=16}

u_r=u_xx_r+u_yy_r+u_zz_r=2xp\cos\theta+zp\sin\theta+y

u_r(2,2,0)=2\cdot4\cdot2\cos0+4\cdot2\sin0+0\implies\boxed{u_r(2,2,0)=16}

u_\theta=u_xx_\theta+u_yy_\theta+u_zz_\theta=-2xpr\sin\theta+zpr\cos\theta

u_\theta(2,2,0)=-2\cdot4\cdot2\cdot2\sin0+4\cdot2\cdot2\cos0\implies\boxed{u_\theta(2,2,0)=16}

7 0
3 years ago
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