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Mathematics

1 answer:

vampirchik [111]3 years ago

5 0

Answer:

third quadrant.

Step-by-step explanation:

The points of Quadrant I have both positive x and y values.

The points of Quadrant II have negative x and positive y values.

The points of Quadrant III have negative x and y values.

The points of Quadrant IV have positive x and negative y values.

Since -2 and -5 are negative the answer is Quadrant III.

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What is 0.69¯¯¯¯ expressed as a fraction in simplest form? Enter your answer in the box.

Ghella [55]

I belive its 3.555 because i took a educational guess lol

6 0

3 years ago

What’s the area of a circle with a diameter of 4

umka21 [38]

Answer:

$4\pi$ ≈ 12.57 units squared

Step-by-step explanation:

The area of a circle is denoted by: $A=\pi r^{2}$ , where r is the radius.

Remember that diameter is twice the radius, so since the diameter equals 4, we know that the radius r = 4/2 = 2.

Now, plug this in: $A=\pi *2^2=4\pi$

We can round 4pi to 12.57 units squared.

Hope this helps!

7 0

3 years ago

Read 2 more answers

Let C be the positively oriented square with vertices (0,0), (1,0), (1,1), (0,1). Use Green's Theorem to evaluate the line integ

liq [111]

Answer:

1/2

Step-by-step explanation:

The interior of the square is the region D = { (x,y) : 0 ≤ x,y ≤1 }. We call L(x,y) = 7y²x, M(x,y) = 8x²y. Since C is positively oriented, Green Theorem states that

$\int\limits_C {L(x,y)} \, dx + {M(x,y)} \, dy = \int\limits^1_0\int\limits^1_0 {(Mx - Ly)} \, dxdy$

Lets calculate the partial derivates of M and L, Mx and Ly. They can be computed by taking the derivate of the respective value, treating the other variable as a constant.

Mx(x,y) = d/dx 8x²y = 16xy
Ly(x,y) = d/dy 7y²x = 14xy

Thus, Mx(x,y) - Ly(x,y) = 2xy, and therefore, the line ntegral is equal to the double integral

$\int\limits^1_0\int\limits^1_0 {2xy} \, dxdy$

We can compute the double integral by applying the Barrow's Rule, a primitive of 2xy under the variable x is x²y, thus the double integral can be computed as follows

$\int\limits^1_0\int\limits^1_0 {2xy} \, dxdy = \int\limits^1_0 {x^2y} |^1_0 \,dy = \int\limits^1_0 {y} \, dy = \frac{y^2}{2} \, |^1_0 = 1/2$