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3 years ago

0.661 rounded to nearest hundretg

1 answer:

8 0

Since you are doing decimals, know that you are mirroring what you would do on the left side

Your decimal number: .661

The first 6, which is .6 is the tenths place

The second 6, which is .06 is the hundredths place

1 which is .001 is the thousands place

The farther you go left, the bigger the digit number value

So when rounding up, you look at the digit after the number you are rounding

So what is after the hundredths place? Its the thousandths place which is 1

Is 61 closer to 60 or 70? Since 1 is smaller than the halfway point which is 5, its closer to 60

Answer: .661 rounded to the nearest hundredths is .660 or just simply written as .66

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amm1812

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4 years ago

One package of socks cost $7. How many packages can you but with $56?

Akimi4 [234]

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Hope it helps!

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3 years ago

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4 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

3 0

3 years ago