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2 years ago

What is the solution to -3x+6

Mathematics

1 answer:

iren [92.7K]2 years ago

3 0

Answer:

Step-by-step explanation:

-3+6=3 then add the x= 3x

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WILL GET BRAINLEST WILL HAVE TI ZOOM IN THI SHIW WORK TOO PLSSS ASAP

Lynna [10]

Answer:

sin = 1.333333

cos = 0.75

8 0

3 years ago

1. Given a right circular conea) What is the shape of a cross section parallel to the base of the cone?b) what is the shape of a

AlekseyPX

Given:

There are given the statement, shape of the cross-section that is parallel to the base of the cone.

Explanation:

If the base of the cone that parallels to tye cross-section.

So,

The shape of the cross-section is

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8 months ago

Find the slope of the given points (6, -10) and (-15, 15). Show your work.

charle [14.2K]

Answer:

Step-by-step explanation:

The slope of a line between two points can be calculated by dividing the difference of y values by the difference of x values

in this case

$m = \frac{y_2-y_1}{x_2-x_1} = \frac{15-(-10)}{-15-6} = \frac{25}{-21} = \frac{-25}{21}$

6 0

2 years ago

Function A

Viktor [21]

The answer would be 9 because

3 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

2 years ago