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Anarel [89]
3 years ago
6

Which ordered pair describes the location of point U?

Mathematics
1 answer:
Pie3 years ago
8 0
Point U is at (2, -3).

I hope this helps!
You might be interested in
A ladder is placed 30 inches from a wall. It touches the wall at a height of 50 inches from the ground.
Aneli [31]
Answer : 59.04 is the angle 
8 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
5(x+3)-5<br> simplify this.
nignag [31]

Answer:

5x+10

Step-by-step explanation:

Distribute:

=(5)(x)+(5)(3)+−5

=5x+15+−5

Combine Like Terms:

=5x+15+−5

=(5x)+(15+−5)

=5x+10

8 0
3 years ago
The number of frogs on the pond is Proportional to the number of salamanders placed in the pond. Suppose there are 10 Frog’s for
bonufazy [111]

Answer:

Y=10x

(multiply 10 times the salamanders)

Step-by-step explanation:

a formula that can be used for this circumstance is a linear y=mx+b problem. for 1 salamander there are 10 frogs. so with this we can state that y=10x. whatever you plug into the x for the number of salamanders will give you the answer of how many frogs are in the pond. for example.

Y=10x

Y=10(3)

Y=30

30 frogs.

also because 10 is a nice number you can just multiply the number of salamanders times 10. 3*10=30

4 0
3 years ago
NO LINKS!!
dexar [7]

Answer: Anything between 0 and 10, excluding both endpoints.

In terms of symbols we can say 0 < w < 10 where w is the width.

===================================================

Explanation:

You could do this with two variables, but I think it's easier to instead use one variable only. This is because the length is dependent on what you pick for the width.

w = width

2w = twice the width

2w-5 = five less than twice the width = length

So,

  • width = w
  • length = 2w-5

which lead to

area = length*width

area = (2w-5)*w

area = 2w^2-5w

area < 150

2w^2 - 5w < 150

2w^2 - 5w - 150 < 0

To solve this inequality, we will solve the equation 2w^2-5w-150 = 0

Use the quadratic formula. Plug in a = 2, b = -5, c = -150

w = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\w = \frac{-(-5)\pm\sqrt{(-5)^2-4(2)(-150)}}{2(2)}\\\\w = \frac{5\pm\sqrt{1225}}{4}\\\\w = \frac{5\pm35}{4}\\\\w = \frac{5+35}{4} \ \text{ or } \ w = \frac{5-35}{4}\\\\w = \frac{40}{4} \ \text{ or } \ w = \frac{-30}{4}\\\\w = 10 \ \text{ or } \ w = -7.5\\\\

Ignore the negative solution as it makes no sense to have a negative width.

The only practical root is w = 10.

If w = 10 feet, then the area = 2w^2-5w results in 150 square feet.

----------------------

Based on that root, we need to try a sample value that is to the left of it.

Let's say we try w = 5.

2w^2 - 5w < 150

2*5^2 - 5*5 < 150

25 < 150 ... which is true

This shows that if 0 < w < 10, then 2w^2-5w < 150 is true.

Now try something to the right of 10. I'll pick w = 15

2w^2 - 5w < 150

2*15^2 - 5*15 < 150

375 < 150 ... which is false

It means w > 10 leads to 2w^2-5w < 150  being false.

Therefore w > 10 isn't allowed if we want 2w^2-5w < 150 to be true.

4 0
2 years ago
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