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3 years ago

Two northern towns are at record low temperatures. One is at -13 degrees and the

Mathematics

1 answer:

marin [14]3 years ago

7 0

Answer:

9 degrees is the answer you are looking for

Step-by-step explanation:

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PLISSSSSS HELP THIS IS FOR EXAM THERE IS A PICTURE

Marizza181 [45]

Answer:

elimination anyday

Step-by-step explanation:

there are two 3y terms so they cancel out when using elimination, mark me brainliest plz

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2 years ago

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The set of all output values y in a function is called the ___ of the function

Debora [2.8K]

Range of a function.

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3 years ago

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Amanda recorded the temperatures at which two oils freeze:

lianna [129]

Answer:

Peanut oil freezes at a higher temperature than olive oil because 3.1 °C > –6.1 °C.

Step-by-step explanation:

Well since 3.1 is greater than -6.1 then it should be Peanut oil freezes at a high temperature. I did the test and I got it right. I hope you do to. :)

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2 years ago

1. Let f(x, y) be a differentiable function in the variables x and y. Let r and θ the polar coordinates,and set g(r, θ) = f(r co

Olenka [21]

Answer:

$g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}\\$

Step-by-step explanation:

First, notice that:

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}cos(\frac{\pi}{4}),\sqrt{2}sin(\frac{\pi}{4}))\\$

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}(\frac{1}{\sqrt{2}}),\sqrt{2}(\frac{1}{\sqrt{2}}))\\$

$g(\sqrt{2},\frac{\pi}{4})=f(1,1)\\$

We proceed to use the chain rule to find $g_{r}(\sqrt{2},\frac{\pi}{4})$ using the fact that $X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)$ to find their derivatives:

$g_{r}(r,\theta)=f_{r}(rcos(\theta),rsin(\theta))=f_{x}( rcos(\theta),rsin(\theta))\frac{\delta x}{\delta r}(r,\theta)+f_{y}(rcos(\theta),rsin(\theta))\frac{\delta y}{\delta r}(r,\theta)\\$

Because we know $X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)$ then:

$\frac{\delta x}{\delta r}=cos(\theta)\ and\ \frac{\delta y}{\delta r}=sin(\theta)$

We substitute in what we had:

$g_{r}(r,\theta)=f_{x}( rcos(\theta),rsin(\theta))cos(\theta)+f_{y}(rcos(\theta),rsin(\theta))sin(\theta)$

Now we put in the values $r=\sqrt{2}\ and\ \theta=\frac{\pi}{4}$ in the formula:

$g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=f_{x}(1,1)cos(\frac{\pi}{4})+f_{y}(1,1)sin(\frac{\pi}{4})$

Because of what we supposed:

$g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=-2cos(\frac{\pi}{4})+3sin(\frac{\pi}{4})$

And we operate to discover that:

$g_{r}(\sqrt{2},\frac{\pi}{4})=-2\frac{\sqrt{2}}{2}+3\frac{\sqrt{2}}{2}$

$g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}$

and this will be our answer

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3 years ago

The table shows how much an air-conditioning repair company charges for different numbers of hours of work. Graph the data, and

emmainna [20.7K]

How someone supposed to know what the table looks like..... at least add the table

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3 years ago

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