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2 years ago

The perimeter of a figure is the total length of all of its sides.

Mathematics

1 answer:

Fantom [35]2 years ago

6 0

Answer:

True.

Step-by-step explanation:

That is exactly what a perimeter is. Hope this helps

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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.

If that's the case, then

$z_s=z_xx_s+z_yy_s$

$z_t=z_xx_t+z_yy_t$

as the hint suggests. We have

$z=\sin x\cos y\implies\begin{cases}z_x=\cos x\cos y=\cos(s+t)\cos(s^2t)\\z_y=-\sin x\sin y=-\sin(s+t)\sin(s^2t)\end{cases}$

$x=s+t\implies x_s=x_t=1$

$y=s^2t\implies\begin{cases}y_s=2st\\y_t=s^2\end{cases}$

Putting everything together, we get

$z_s=\cos(s+t)\cos(s^2t)-2st\sin(s+t)\sin(s^2t)$

$z_t=\cos(s+t)\cos(s^2t)-s^2\sin(s+t)\sin(s^2t)$

8 0

3 years ago

An education researcher claims that 58% of college students work year-round. In a random sample of 400 college students, 232

Hatshy [7]

Answer:

The proportion of college students who work year-round is 58%.

Step-by-step explanation:

The claim made by the education researcher is that 58% of college students work year-round.

A random sample of 400 college students, 232 say they work year-round.

To test the researcher's claim use a one-proportion z-test.

The hypothesis can be defined as follows:

H₀: The proportion of college students who work year-round is 58%, i.e. p = 0.58.

Hₐ: The proportion of college students who work year-round is 58%, i.e. p ≠ 0.58. C

Compute the sample proportion as follows:

$\hat p=\frac{232}{400}=0.58$

Compute the test statistic value as follows:

$z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.58-0.58}{\sqrt{\frac{0.58(1-0.58)}{400}}}=0$

The test statistic value is 0.

Decision rule:

If the p-value of the test is less than the significance level then the null hypothesis will be rejected.

Compute the p-value for the two-tailed test as follows:

$p-value=2\times P(z$

*Use a z-table for the probability.

The p-value of the test is 1.

The p-value of the test is very large when compared to the significance level.

The null hypothesis will not be rejected.

Thus, it can be concluded that the proportion of college students who work year-round is 58%.

6 0

3 years ago

Simplify using the laws of exponents (4^3)^-2 × (2^3)^4 ×(8/15)^-2

katrin2010 [14]

Answer:

$\dfrac{225}{64}$

Step-by-step explanation:

$(4^3)^{-2} \times (2^3)^4 \times (\dfrac{8}{15})^{-2} =$

$= (2^2)^{-6} \times 2^{12} \times (\dfrac{15}{8})^{2}$

$= 2^{-12} \times 2^{12} \times \dfrac{225}{64}$

$= \dfrac{225}{64}$

8 0

3 years ago

Max is stacking logs at his campground for firewood. After his first load of logs, he has 12 logs on the stack. After his sevent

Natasha2012 [34]

Answer:

Let L(n) be the function that gives the amount of logs stacked after n loads.

L(n) = 12 + 8(n-1)

Step-by-step explanation:

Let L(n) be the function that gives the amount of logs stacked after n loads.

Let's call for the moment the first load as L(0)

L(0)= 12

Let r be the number of logs carried in each load, then

L(n) = 12 + nr

Since L(6) (the seventh load) equals 60, we have

60 = 12 + 6r, and r = 8.

So a function for the number of loads starting from n=0 would be

L(n) = 12 + 8n

If we want to start with n=1, we simply change the variable

L(n) = 12 + 8(n-1) (n=1,2,3,...).

So L(1) = 12, L(2) = 20, L(3) = 28,...L(7) = 60 and so on.

7 0

3 years ago

C=3 c2 + 5 ..............

vaieri [72.5K]

Answer:

3x2+5=11

Step-by-step explanation:

8 0

2 years ago