??? Need help please

In the given right triangle, find the length of the hypotenuse, AB, if BC = 6 and AC = 5.

wolverine [178]

*Use the Pythagorean Theorem

a^2 + b^2 = c^2

6^2 + 5^2 - c^2
36 + 25 = c^2
61 = c^2 *Square root 61

c^2 (c squared) = about 7.8

6 0

3 years ago

EASY QUESTION 100 Points Topic: Multiplying and dividing fractions

Rufina [12.5K]

Answer:

4 times as fast

Step-by-step explanation:

To determine how many time as fast, take the rate up and divide by the rate down

63 minutes

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

15 minute 45 seconds

But we need to be all in the same units so change second to minutes by dividing by 60

63 minutes

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

15 minute 45/60 minutes

63 minutes

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

15.75 minute

4 times

8 0

3 years ago

Read 2 more answers

An ice cream shop sells two type of cones: cake cones for $1.75 and waffle cones for $3.25. on tuesday, they sold 50 more waffle

Nezavi [6.7K]

Answer:

sorry no idea

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

line k has a slope of -5. line j is perpendicular to line k and passes though the point (5,9) create the equation for line j

Anna71 [15]

Answer:

y = 1/5x + 8

Step-by-step explanation:

The first thing you should know is that the slope of a perpendicular line is the opposite reciprocal. So, the slope of line j is 1/5. This is your current equation: y=1/5x + b. Plug in your coordinate to the equation, it should look like this: 9=1/5(5)+ b. 1/5 times 5 equals 1, so this is what it should look like: 9=1 + b. Subtract 1 from both sides to get 8 = b. Go back to your original equation and plug in 8 to b. This is your final equation: y = 1/5x + 8. Hope this helped!

3 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

￼??? Need help please

??? Need help please