On a number line 8.63 would be located where

Mathematics

1 answer:

Oksanka [162]3 years ago

4 0

Answer:

It would be slightly higher than 8.5 and lower than 9.

Step-by-step explanation:

8.6 is bigger than 8.5, but it's smaller than 9, so you would place it in between the two but closer to the 8.5 mark.

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Let φ(x, y) = arctan (y/x) .

Alexandra [31]

Answer:

a) $\large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})$

b) $\large \mathbb{R}^2-\{(0,0)\}$

c) the points of the form (x, -x) for x≠0

Step-by-step explanation:

If φ(x, y) = arctan (y/x), the vector field F = ∇φ would be

$\large F(x,y)=(\frac{\partial \phi}{\partial x},\frac{\partial \phi}{\partial y})$

On one hand we have,

$\large \frac{\partial \phi}{\partial x}=\frac{\partial arctan(y/x)}{\partial x}=\frac{-y/x^2}{1+(y/x)^2}=-\frac{y/x^2}{1+y^2/x^2}=\\\\=-\frac{y/x^2}{(x^2+y^2)/x^2}=-\frac{y}{x^2+y^2}$

On the other hand,

$\large \frac{\partial \phi}{\partial y}=\frac{\partial arctan(y/x)}{\partial y}=\frac{1/x}{1+(y/x)^2}=\frac{1/x}{1+y^2/x^2}=\\\\=\frac{1/x}{(x^2+y^2)/x^2}=\frac{x}{x^2+y^2}$

$\large F(x,y)=(-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})$

The domain of definition of F is

$\large \mathbb{R}^2-\{(0,0)\}$

i.e., all the plane X-Y except the (0,0)

Here we want to find all the points such that

$\large (-\frac{y}{x^2+y^2},\frac{x}{x^2+y^2})=(k,k)$

where k is a real number other than 0.

But this means

$\large -\frac{y}{x^2+y^2}=\frac{x}{x^2+y^2}\Rightarrow y=-x$

So, all the points in the line y = -x except (0,0) are parallel to the vector field F, that is, the points (x, -x) with x≠ 0

8 0

3 years ago

ASAP HELP! BRAINLIEST! can someone help me with this? i need work shown. will mark brainliest!

ki77a [65]

Answer:

4x^2 +12x +44 remainder 161x +84

Step-by-step explanation:

At each step, the quotient term is the ratio of the leading dividend term to the leading divisor term. The first quotient term, for example, is ...

(4x^4)/(x^2) = 4x^2

The quotient term found this way is multiplied by the divisor and subtracted from the dividend. The difference is the new dividend and the process repeats.

You're done when the degree of the dividend is less than the degree of the divisor. This remainder can be expressed as a fraction with the divisor as the denominator.

$\dfrac{4x^4+5x-4}{x^2-3x-2}=4x^2+12x+44+\dfrac{161x+84}{x^2-3x-2}$