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Tcecarenko [31]
3 years ago
8

No need to show your work.

Mathematics
2 answers:
tino4ka555 [31]3 years ago
8 0

Answer:

Hello! I dont know for sure but I think the answer is 7 2/5 answer at your own risk though.

Hope that helps!

Delvig [45]3 years ago
7 0

Answer:

7 2/5

Step-by-step explanation:

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can you mark me of a brainliest please answers to you is C

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What number is greater 17 or 20/5
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17 is greater than 20/5 because 20 divided by 5=4
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7x/2+5=8<br><br> solve for x
ANEK [815]

Answer:

X= 6/7

Step-by-step explanation:

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Terry earns a base salary of $2100 per month. Once he reaches $40000 in total sales, he earns an additional 5% commission (salar
Zigmanuir [339]

Answer:

<h2>$4200 </h2>

Step-by-step explanation:

Step one:

given data

we are told that Terry's base salary is $2100  per month.

in addition 5% on sales greater than $40000.

let the monthly salary be M(s)

and let "s" be the amount of sales made

his monthly commission =0.05s

Step two:

The expression for his monthly earning is given as

M(s)=0.05s+2100   where s>40000

 hence for a month when s=42000

M(s)=0.05(42000)+2100

M(s)=2100+2100

M(s)=4200

<u>Therefore for an instance when s=42000 his earnings  for the month will be $4200 </u>

7 0
3 years ago
Compute the matrix of partial derivatives of the following functions.
s344n2d4d5 [400]

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}

(c)

Df(x,y)=\begin{bmatrix}y&x\\1&-1\\y&x\end{bmatrix}

(d)

Df(x,y,z)=\begin{bmatrix}1&0&1\\0&1&0\\1&-1&0\end{bmatrix}

5 0
3 years ago
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