Find the total cost.

Find the domaine of y= square root x+6

babunello [35]

Two is the domain of them

5 0

3 years ago

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$45.00 for 9 pounds what is the unit rate?

dmitriy555 [2]

$45/9 pounds
$5 per pound

7 0

3 years ago

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A club with five members will randomly select the president and vice president. Each possible pair will be assigned an equal por

Tju [1.3M]

Answer:

C. 20

Step-by-step explanation:

The order in which the members are selected is important. The first one selected is the president and the second is the vice president.

So we use the permutations formula to solve this question.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

$P_{(n,x)} = \frac{n!}{(n-x)!}$

In this problem:

2 people are going to be chosen from a set of 5. So

$P_{(5,2)} = \frac{5!}{(5-2)!} = 20$

So the correct answer is:

C. 20

3 0

3 years ago

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What is the value of x in the equation 8x-5(x-1)=20 <br> 1. 25/3<br> 2. 21/4<br> 3. 5<br> 4. 7

Ede4ka [16]

Answer : x = 5
So pick “(3)”

8 0

3 years ago

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:

a)

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,

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

So

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

b)

The domain of definition of F is

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

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

c)

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