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3 years ago

You owe 1200 on your credit card which charges 1.5 per month

Mathematics

1 answer:

Ivahew [28]3 years ago

6 0

40 meses com um custo de 4030 reais por mes espero ter ajudado bjjs

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If, for each mouse, the probability of turning left is 0.5 and the probability of turning right is 0.5, what is the probability

Phoenix [80]

Answer:

.125

Step-by-step explanation:

(.5)^2(.5)^1= .125

Where (.5)^2 is the probability two mice turn left multiplied by (.5) that one of them turns right.

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3 years ago

Luis was served 145 grams of meat and 217 grams of vegtables at a meal what was the total mass of the meat and the vegtables ?

AlladinOne [14]

Hi! Since it is asking for the total, you must add!

145+217=?

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3 years ago

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How can I work this out or solve this?

Ede4ka [16]

Add it all up, the answer would be B. 2.37

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3 years ago

Geometry math question no Guessing and Please show work

Andru [333]

D. (7,0)
The original coordinates were (0,7)y and (1,5)x then I rotated it 90 degrees and the coordinates were (7,0)y and (5,1)x

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3 years ago

Use the Chain Rule (Calculus 2)

atroni [7]

1. By the chain rule,

$\dfrac{\mathrm dz}{\mathrm dt}=\dfrac{\partial z}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial z}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}$

I'm going to switch up the notation to save space, so for example, $z_x$ is shorthand for $\frac{\partial z}{\partial x}$ .

$z_t=z_xx_t+z_yy_t$

We have

$x=e^{-t}\implies x_t=-e^{-t}$

$y=e^t\implies y_t=e^t$

$z=\tan(xy)\implies\begin{cases}z_x=y\sec^2(xy)=e^t\sec^2(1)\\z_y=x\sec^2(xy)=e^{-t}\sec^2(1)\end{cases}$

$\implies z_t=e^t\sec^2(1)(-e^{-t})+e^{-t}\sec^2(1)e^t=0$

Similarly,

$w_t=w_xx_t+w_yy_t+w_zz_t$

where

$x=\cosh^2t\implies x_t=2\cosh t\sinh t$

$y=\sinh^2t\implies y_t=2\cosh t\sinh t$

$z=t\implies z_t=1$

To capture all the partial derivatives of $w$ , compute its gradient:

$\nabla w=\langle w_x,w_y,w_z\rangle=\dfrac{\langle1,-1,1\rangle}{\sqrt{1-(x-y+z)^2}}}=\dfrac{\langle1,-1,1\rangle}{\sqrt{-2t-t^2}}$

$\implies w_t=\dfrac1{\sqrt{-2t-t^2}}$

2. The problem is asking for $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ . But $z$ is already a function of $x,y$ , so the chain rule isn't needed here. I suspect it's supposed to say "find $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}$ " instead.