What is the domain of the square root function graphed below? x=-1 x>/=-1

Mathematics

1 answer:

liq [111]3 years ago

8 0

Answer:

im guessing -1

Step-by-step explain

well it would be greater than -1x-1

You might be interested in

What is the value of x?<br><br> x = ___

Shalnov [3]

Answer:

x = 24

Step-by-step explanation:

The triangle is split into two similar triangles. They have the same shape but not the same size. To solve, write a proportion.

$\frac{56}{44.8}=\frac{35}{x+4}$

Cross multiply to solve for x.

56(x+4) = 44.8(35)

56x + 224 = 1,568

56x = 1,344

x= 24

4 0

3 years ago

Whats the answer?please tell me!!

zavuch27 [327]

95.72 + 5.228 = 100.948

6 0

3 years ago

Read 2 more answers

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,

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