Using the protractor, find the angle measure.

0.8, 1.2, 1.6, 2.0, ... what is the rule in the squennce above?

belka [17]

Answer:

You add 0.4 each time

Step-by-step explanation:

.........

8 0

3 years ago

Read 2 more answers

What is the length of BC in the right triangle below?

telo118 [61]

Answer:

The answer should be 15

Step-by-step explanation:

5 0

3 years ago

Use a graphing calculator to approximate the vertex of the graph of the parabola defined by the following

vivado [14]

Answer: (-0.4, 6.8)

The graph is attached

Step-by-step explanation:

We can find the axis of symmetry using the formula x = -b/2a

x = - -4/2(-5)

x = 4/-10

x = 0.04

Substitute for x and calculate y

y = $\left(0.4^{2}\right)\left(-5\right)-4\left(-0.4\right)\ +6$

y = 6.8

3 0

3 years ago

Find the value of MN if AB= 21 cm, BC= 16.8 cm, and LM= 28 cm.

Tom [10]

Answer:

lm=36x2

Step-by-step explanation:

8 0

3 years ago

Find the counterclockwise circulation and outward flux of the field F=7xyi+5y^2j around and over the boundary of the region C en

dezoksy [38]

Split up the boundary of C (which I denote ∂C throughout) into the parabolic segment from (1, 1) to (0, 0) (the part corresponding to y = x ²), and the line segment from (1, 1) to (0, 0) (the part of ∂C on the line y = x).

Parameterize these pieces respectively by

r(t) = x(t) i + y(t) j = t i + t ² j

and

s(t) = x(t) i + y(t) j = (1 - t ) i + (1 - t ) j

both with 0 ≤ t ≤ 1.

The circulation of F around ∂C is given by the line integral with respect to arc length,

$\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf T \,\mathrm ds$

where T denotes the tangent vector to ∂C. Split up the integral over each piece of ∂C :

• on the parabolic segment, we have

T = dr/dt = i + 2t j

• on the line segment,

T = ds/dt = -i - j

Then the circulation is

$\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf T\,\mathrm ds = \int_0^1 (7t^3\,\mathbf i+5t^4\,\mathbf j)\cdot(\mathbf i+2t\,\mathbf j)\,\mathrm dt + \int_0^1 (7(1-t)^2\,\mathbf i+5(1-t)^2\,\mathbf j)\cdot(-\mathbf i-\mathbf j)\,\mathrm dt \\\\ = \int_0^1 (7t^3+10t^5)\,\mathrm dt - 12 \int_0^1 (1-t)^2\,\mathrm dt =\boxed{-\frac7{12}}$

Alternatively, we can use Green's theorem to compute the circulation, as

$\displaystyle\int_{\partial C}\mathbf F\cdot\mathbf T\,\mathrm ds = \iint_C\frac{\partial(5y^2)}{\partial x} - \frac{\partial(7xy)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = -7\int_0^1\int_{x^2}^x x\,\mathrm dx \\\\ = -7\int_0^1 xy\bigg|_{y=x^2}^{y=x}\,\mathrm dx \\\\ =-7\int_0^1(x^2-x^3)\,\mathrm dx = -\frac7{12}$

The flux of F across ∂C is

$\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf N \,\mathrm ds$

where N is the normal vector to ∂C. While T = x'(t) i + y'(t) j, the normal vector is N = y'(t) i - x'(t) j.

• on the parabolic segment,

N = 2t i - j

• on the line segment,

N = - i + j

So the flux is

$\displaystyle \int_{\partial C}\mathbf F\cdot\mathbf N\,\mathrm ds = \int_0^1 (7t^3\,\mathbf i+5t^4\,\mathbf j)\cdot(2t\,\mathbf i-\mathbf j)\,\mathrm dt + \int_0^1 (7(1-t)^2\,\mathbf i+5(1-t)^2\,\mathbf j)\cdot(-\mathbf i+\mathbf j)\,\mathrm dt \\\\ = \int_0^1 (14t^4-5t^4)\,\mathrm dt - 2 \int_0^1 (1-t)^2\,\mathrm dt =\boxed{\frac{17}{15}}$

5 0

3 years ago