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Darina [25.2K]
3 years ago
6

Y=5/9x+9, find the slope

Mathematics
2 answers:
sashaice [31]3 years ago
5 0

Answer:

slope is 5 / 9 .

................................

Igoryamba3 years ago
3 0
the slope would be 4/9
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Determine whether the following function is a​ one-to-one function. If it is​ one-to-one, list the inverse function by switching
BartSMP [9]

Answer: An easy way to determine whether a function is a one-to-one function is to use the horizontal line test on the graph of the function. To do this, draw horizontal lines through the graph. If any horizontal line intersects the graph more than once, then the graph does not represent a one-to-one function.

i dont know the exact answer but its a lil sum

Step-by-step explanation:

We know that for any two inverses f(x) = g(y), meaning that if we take f(x) for any x in the domain of f(x), then g(y), where y is the outcome of f(x), should output x. So that is a simple test to see if two functions are inverses.

8 0
3 years ago
Plot the following points to create a rectangle. Find the missing vertex. (4,0);(-6,0);(-6,-4)
Sveta_85 [38]

Answer:

(4,-4)

Step-by-step explanation:

7 0
3 years ago
Please help i cannot get it ​
Lerok [7]
Pi symbol raised to the 3 times x raised to the 4
7 0
3 years ago
Let F⃗ =2(x+y)i⃗ +8sin(y)j⃗ .
Alik [6]

Answer:

-42

Step-by-step explanation:

The objective is to find the line integral of F around the perimeter of the rectangle with corners (4,0), (4,3), (−3,3), (−3,0), traversed in that order.

We will use <em>the Green's Theorem </em>to evaluate this integral. The rectangle is presented below.

We have that

           F(x,y) = 2(x+y)i + 8j \sin y = \langle 2(x+y), 8\sin y \rangle

Therefore,

                  P(x,y) = 2(x+y) \quad \wedge \quad Q(x,y) = 8\sin y

Let's calculate the needed partial derivatives.

                              P_y = \frac{\partial P}{\partial y} (x,y) = (2(x+y))'_y = 2\\Q_x =\frac{\partial Q}{\partial x} (x,y) = (8\sin y)'_x = 0

Thus,

                                    Q_x -P_y = 0 -2 = - 2

Now, by the Green's theorem, we have

\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA = \int \limits_{-3}^{4} \int \limits_{0}^{3} (-2)\,dy\, dx \\ \\\phantom{\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA}= \int \limits_{-3}^{4} (-2y) \Big|_{0}^{3} \; dx\\ \phantom{\oint_C F \,dr = \iint_D (Q_x-P_y)\,dA}= \int \limits_{-3}^{4} (-6)\; dx = -6x  \Big|_{-3}^{4} = -42

4 0
3 years ago
What is the equation ?
andreev551 [17]

Answer:

y=x-5

Step-by-step explanation:

y-2=(7-2)/(12-7)(x-7)

y=1(x-7)+2

y=x-5

3 0
3 years ago
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