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3 years ago

Solve the M -5 m = -35

Mathematics

1 answer:

dedylja [7]3 years ago

4 0

Answer: m = 7

Step-by-step explanation: In this equation, since a -5 is being multiplied by m, in order to get m by itself, we must divide both sides of the equation by -5.

On the left side, our -5's cancel out and we are left with m. On the right side, -35 ÷ -5 gives us 7. So m = 7.

To check our answer, we plug 7 back in for m in the original equation and we get -5 (7) = -35 which is a true statement so we know our answer is correct.

Don't just do this problem in your head. It's extremely important to develop the habit of putting all your steps down on paper or digitally. It will really pay off for you down the line.

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Complete the set of ordered pairs for the relation.

neonofarm [45]

The set of ordered pairs is:

{(-2, 2), (-1, 0), (0, 2), (1, 4), (2, 6)}

<h3>How to complete the set of ordered pairs?</h3>

Here we have the relation:

y = 2*|x + 1|

And the domain is x ∈ { -2, -1, 0, 1, 2}

Evaluating in the values of the domain, we get:

if x = -2.

y = 2*|-2 + 1| = 2

Then we have the ordered pair (-2, 2)

if x = -1

y = 2*|-1 + 1| = 0

Then we have the ordered pair (-1, 0)

if x = 0

y = 2*|0 + 1| = 2

Then we have the ordered pair (0, 2)

if x = 1

y = 2*|1 + 1| = 4

Then we have the ordered pair (1, 4)

if x = 2

y = 2*|2 + 1| = 6

Then we have the ordered pair (2, 6)

The set of ordered pairs is:

{(-2, 2), (-1, 0), (0, 2), (1, 4), (2, 6)}

If you want to learn more about ordered pairs:

brainly.com/question/1528681

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6 0

1 year 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 the Green's Theorem 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

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