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3 years ago

Natalia is writing a recursiye formula to represent the

Mathematics

2 answers:

kolbaska11 [484]3 years ago

7 0

Answer:

3/2

Step-by-step explanation:

3/2 is the value

Nadya [2.5K]3 years ago

4 0

Answer:

1.5

Step-by-step explanation:

As the common difference is not same for all terms, the given sequence is a geometric sequence

the standard formula for a geometric sequence is:

$a_n=a_1r^{n-1}$

The formula to calculate common ratio is:

$r = \frac{a_n}{a_{n-1}}$

Dividing the term by previous term

So,

r = 12/8 = 18/12 = 27/18 = 1.5

The value for common ratio will be:

1.5 ..

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What’s the answer to this???

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Answer:

5+6-8

Step-by-step explanation:

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2 years ago

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24 x 10 x 90 = 90 x 2,400 True or False?

xenn [34]

False, 24 x 10 x 90 = 21,600, 90 x 2,400 = 216,000

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3 years ago

Read 2 more answers

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.