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3 years ago

Find the Area of the Trapezoid.

Mathematics

1 answer:

AfilCa [17]3 years ago

6 0

Answer:

18 ft^2

Step-by-step explanation:

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Find the length of the arc of a circle with radius of 20 centimeters intercepted by a central angle of 45 degrees

oksian1 [2.3K]

If the radius is 20cm, the circumference is

$C = 2\pi r = 40\pi$ centimeters.

Now, 45° is one eighth of a whole turn, so the arc intercepted by a 45° angle is one eighth of the whole circumference, so the answer is

$\dfrac{40\pi}{8} = 5\pi$

8 0

3 years ago

5. given the vector [-5,12], find each of the following vectors. a. same direction, twice as long. b. same direction, length 1 (

pshichka [43]

The vectors are listed below:

v = [- 10 / 13, 24 / 13]
v = [- 5 / 13, 12 / 13]
v = [50 / 13, - 120 / 13]

<h3>How to derive the expressions of vectors</h3>

Vectors are characterized by magnitude and direction. The magnitude is a scalar and the direction by a ordered pair, whose formula is shown below:

v = k · u

Where u is a unit vector, that is, a vector of length 1.

Now we proceed to determine each vector:

a) The equation of the vector:

r = √[(- 5)² + 12²]

r = 13

k = 2

v = (2 / 13) · [- 5, 12]

v = [- 10 / 13, 24 / 13]

b) The equation of the vector:

v = (1 / 13) · [- 5, 12]

v = [- 5 / 13, 12 / 13]

c) The equation of the vector:

v = (- 10 / 13) · [- 5, 12]

v = [50 / 13, - 120 / 13]

To learn more on vectors: brainly.com/question/4579006

#SPJ1

5 0

2 years ago

who wants to be my math tutor for rn plz? i'll give u extra points if u help me for awhile? whos willing to do it?? like actuall

Alex

I would help, but my math isn’t even that good

8 0

3 years ago

For each vector field f⃗ (x,y,z), compute the curl of f⃗ and, if possible, find a function f(x,y,z) so that f⃗ =∇f. if no such f

butalik [34]

$\vec f(x,y,z)=(2yze^{2xyz}+4z^2\cos(xz^2))\,\vec\imath+2xze^{2xyz}\,\vec\jmath+(2xye^{2xyz}+8xz\cos(xz^2))\,\vec k$

Let

$\vec f=f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k$

The curl is

$\nabla\cdot\vec f=(\partial_x\,\vec\imath+\partial_y\,\vec\jmath+\partial_z\,\vec k)\times(f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k)$

where $\partial_\xi$ denotes the partial derivative operator with respect to $\xi$ . Recall that

$\vec\imath\times\vec\jmath=\vec k$

$\vec\jmath\times\vec k=\vec i$

$\vec k\times\vec\imath=\vec\jmath$

and that for any two vectors $\vec a$ and $\vec b$ , $\vec a\times\vec b=-\vec b\times\vec a$ , and $\vec a\times\vec a=\vec0$ .

The cross product reduces to

$\nabla\times\vec f=(\partial_yf_3-\partial_zf_2)\,\vec\imath+(\partial_xf_3-\partial_zf_1)\,\vec\jmath+(\partial_xf_2-\partial_yf_1)\,\vec k$

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

$\nabla\times\vec f=\vec0$

which means $\vec f$ is indeed conservative and we can find $f$ .

Integrate both sides of

$\dfrac{\partial f}{\partial y}=2xze^{2xyz}$

with respect to $y$ and

$\implies f(x,y,z)=e^{2xyz}+g(x,z)$

Differentiate both sides with respect to $x$ and

$\dfrac{\partial f}{\partial x}=\dfrac{\partial(e^{2xyz})}{\partial x}+\dfrac{\partial g}{\partial x}$

$2yze^{2xyz}+4z^2\cos(xz^2)=2yze^{2xyz}+\dfrac{\partial g}{\partial x}$

$4z^2\cos(xz^2)=\dfrac{\partial g}{\partial x}$

$\implies g(x,z)=4\sin(xz^2)+h(z)$

Now

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+h(z)$

and differentiating with respect to $z$ gives

$\dfrac{\partial f}{\partial z}=\dfrac{\partial(e^{2xyz}+4\sin(xz^2))}{\partial z}+\dfrac{\mathrm dh}{\mathrm dz}$

$2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}$

$\dfrac{\mathrm dh}{\mathrm dz}=0$

$\implies h(z)=C$

for some constant $C$ . So

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+C$

3 0

4 years ago

A rectangular napkin has an area of 88 square centimeters and a length of 8 centimeters.

Svetllana [295]

11 centimeters
This is because a rectangle has an area of length times width and so you can divide 88 by 8 to get 11

3 0

3 years ago

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