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Fofino [41]
3 years ago
13

Find the angle measures of a triangle if the first angle is thirty-one degrees more than the second, and the third is five degre

es less than twice the first. Show the equation as well as the complete answer. (Hint: The sum of the interior angles of a triangle is 180 degrees)
Mathematics
1 answer:
MAVERICK [17]3 years ago
6 0

Answer:

23°, 54°, 103°

Step-by-step explanation:

let the2nd angle = x

1st angle = x + 31

3rd angle = 2(x + 31) - 5 = 2x + 62 - 5 = 2x + 57

Sum the 3 angles and equate to 180

x + 31 + x + 2x + 57 = 180 , that is

4x + 88 = 180 ( subtract 88 from both sides )

4x = 92 ( divide both sides by 4 )

x = 23

Then

1st angle = x + 31 = 23 + 31 = 54°

2nd angle = x = 23°

3rd angle = 2x + 57 = 2(23) + 57 = 46 + 57 = 103°

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3/10ths of the money

Step-by-step explanation:

Add together the two numbers to get the total.

Josh gets 30 percent and Lucy gets 70 percent.

3/10

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Step-by-step explanation:

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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}

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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

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

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