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3 years ago

PLEASE HELP I WILL MARK YOU BRAINIEST PLEASE

Mathematics

1 answer:

cupoosta [38]3 years ago

8 0

Answer:

Zeniab makes the most money by $0.25, and would make $2.50 more.

Step-by-step explanation:

Okay, the good news is, this problem is about finding how much you make in an hour.

Zainab makes $44 for 8 hours

and

Keziah makes $47.25 in 9 hours.

in each problem, we can set them up in an equation of

$44=8x

and

$47.25=9x

In the first equation, divide both sides by 8 to isolate the x.

that would be 44/8 = x which is $5.50 per hour for Zainab.

In the second equation, divide both sides by 9 to isolate the x.

That would be 47.25/9 = x which is $5.25 per hour for Keziah.

This means that Zainab would make more per hour, and to find the second part of the question for a 10 hour job,

(You can multiply any number by 10 easily by moving each number up one place. Hundreds go to the thousands when you multiply by 10 as an example. In this problem, 5.50 times 10 would be $55.00 because the 5 behind the decimal would be moved up infront of the decimal)

Zainab would make $55 for 10 hours

Keziah would make $52.50 for 10 hours

Then, take the difference of those final totals, which is $2.50, or multiply $0.25 by 10 to get $2.50

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Step-by-step explanation:

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3 years ago

Compute the matrix of partial derivatives of the following functions.

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For a vector-valued function

$\mathbf f(\mathbf x)=\mathbf f(x_1,x_2,\ldots,x_n)=(f_1(x_1,x_2,\ldots,x_n),\ldots,f_m(x_1,x_2,\ldots,x_n))$

the matrix of partial derivatives (a.k.a. the Jacobian) is the $m\times n$ matrix in which the $(i,j)$ -th entry is the derivative of $f_i$ with respect to $x_j$ :

$D\mathbf f(\mathbf x)=\begin{bmatrix}\dfrac{\partial f_1}{\partial x_1}&\dfrac{\partial f_1}{\partial x_2}&\cdots&\dfrac{\partial f_1}{\partial x_n}\\\dfrac{\partial f_2}{\partial x_1}&\dfrac{\partial f_2}{\partial x_2}&\cdots&\dfrac{\partial f_2}{\partial x_n}\\\vdots&\vdots&\ddots&\vdots\\\dfrac{\partial f_m}{\partial x_1}&\dfrac{\partial f_m}{\partial x_2}&\cdots&\dfrac{\partial f_n}{\partial x_n}\end{bmatrix}$

So we have

(a)

$D f(x,y)=\begin{bmatrix}\dfrac{\partial(e^x)}{\partial x}&\dfrac{\partial(e^x)}{\partial y}\\\dfrac{\partial(\sin(xy))}{\partial x}&\dfrac{\partial(\sin(xy))}{\partial y}\end{bmatrix}=\begin{bmatrix}e^x&0\\y\cos(xy)&x\cos(xy)\end{bmatrix}$

(b)

$D f(x,y,z)=\begin{bmatrix}\dfrac{\partial(x-y)}{\partial x}&\dfrac{\partial(x-y)}{\partial y}&\dfrac{\partial(x-y)}{\partial z}\\\dfrac{\partial(y+z)}{\partial x}&\dfrac{\partial(y+z)}{\partial y}&\dfrac{\partial(y+z)}{\partial z}\end{bmatrix}=\begin{bmatrix}1&-1&0\\0&1&1\end{bmatrix}$