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2 years ago

WRITE a Linear FUNCTION, f, with the given values: f(-4) = -5 and f(2) = -3

Mathematics

1 answer:

Crazy boy [7]2 years ago

8 0

Answer:

answer to the question is A

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Help and I'll give you a McDouble(;

PilotLPTM [1.2K]

Karen sells 7 hot dogs per hour while Henry sells 14 hot dogs .

<span>The rate that Henry sells hot dogs is double the rate that Karen sells corn dogs.</span>

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3 years ago

What is the value of y in the solution to the system of equations?

cluponka [151]

Plug into calculator...
solve(2(-y+1)-3y=-30,y)
y=6.4

7 0

3 years ago

How to solve this problem?

valentinak56 [21]

Very nice handwriting but the math and English are confusing.

Let's assume we're told

$\displaystyle 45 = \sum_{i=1}^9 (x_i - 10)^2$

The subscript is important.

I think we're told the similar sum with 11 gives the smallest possible value for the sum. This is a rather cagey way of telling us 11 is the mean of the nine points. The mean is the number which minimizes the sum of squared deviations.

$\displaystyle 45 = \sum_{i=1}^9 (x_i - 10)^2 = \sum x_i^2 - 20 \sum x_i + 9(100)$

$\displaystyle \sum x_i^2= 20 \sum x_i - 900$

If 11 is the mean, the sum of the points is 9(11)=99.

$\displaystyle \sum x_i^2= 20 (99) - 900 = 1080$

Answer: 1080

6 0

3 years ago

F(x, y, z) = yzexzi + exzj + xyexzk,

Ghella [55]

$\nabla f(x,y,z)=\mathbf F(x,y,z)\iff\dfrac{\partial f}{\partial x}\,\mathbf i+\dfrac{\partial f}{\partial y}\,\mathbf j+\dfrac{\partial f}{\partial z}\,\mathbf k=yze^{xz}\,\mathbf i+e^{xz}\,\mathbf j+xye^{xz}\,\mathbf k$

$\dfrac{\partial f}{\partial x}=yze^{xz}$
$\implies f(x,y,z)=\dfrac{yz}ze^{xz}+g(y,z)=ye^{xz}+g(y,z)$

$\dfrac{\partial f}{\partial y}=e^{xz}=e^{xz}+\dfrac{\partial g}{\partial y}$
$\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)$

$\dfrac{\partial f}{\partial z}=xye^{xz}=xye^{xz}+\dfrac{\mathrm dh}{\mathrm dz}$
$\implies\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=C$

$f(x,y,z)=ye^{xz}+C$

$\mathbf r(t)=(t^2+5)\,\mathbf i+(t^2-1)\,\mathbf j+(t^2-5)\,\mathbf k$
$\implies\mathbf r(0)=5\,\mathbf i-\mathbf j-5\,\mathbf k$
$\implies\mathbf r(5)=30\,\mathbf i+24\,\mathbf j+20\,\mathbf k$

By the gradient theorem, we have for any path $\mathcal C$ originating at the point (5, -1, -5) and terminating at the point (30, 24, 20),

$\displaystyle\int_{\mathcal C}\mathbf F\cdot\mathrm d\mathbf r=f(\mathbf r(5))-f(\mathbf r(0))=f(30,24,20)-f(5,-1,-5)$
$\implies\displaystyle\int_{\mathcal C}\mathbf F\cdot\mathrm d\mathbf r=24e^{600}+e^{-25}$

4 0

3 years ago

Can you write nine proper fraction and nine improper fractions using 3, 5, and 8

ra1l [238]

5/3 8/3 8/5... I don’t think it is possible

6 0

3 years ago