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4 years ago

HALPPPP. Awarding Brainliest!! Pythagorean Theorem.

Mathematics

2 answers:

Novay_Z [31]4 years ago

6 0

Answer:

$\sqrt{29}$

Step-by-step explanation:

Using pythagorean theorem you will get 5 squared plus 2 squared equals c squared. Which is 25+4= 29 = c squared. That's the answer it wont simplify any further.

Verdich [7]4 years ago

5 0

Answer:

d = 5.4

Step-by-step explanation:

Pythagorean theorem: a^2 + b^2 = c^2

a = 2

b = 5

2^2 + 5^2 = c^2

4 + 25 = c^2

29 = c^2

c = 5.4

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*EXTRA PTS* pls answer all 3 :)

dezoksy [38]

Answer:

5, x = 0, y = -2

6, x = 4, Y = 9

7, X = -2, Y = -2

Step-by-step explanation:

8 0

3 years ago

What will be the result of substituting 2 for x in both expressions below? One-half x + 4 x + 6 minus one-half x minus 2

Monica [59]

Answer:

Step-by-step explanation:

The equation is:

1/2x+4x+6-1/2x-2

Ok so you can cancel out 1/2x and 1/2x because 1/2x-1/2x=0

You are left with:

4x+4

If you plug in 2 for x

4(2)+4=12

I hope this helps :)

5 0

3 years ago

Read 2 more answers

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

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