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NeTakaya
3 years ago
12

30 points and brinliest answer! A group of friends go to a family fun park. The park has games and rides.

Mathematics
2 answers:
Ilia_Sergeevich [38]3 years ago
8 0
The person above is correct.
svet-max [94.6K]3 years ago
4 0

Answer:

It costs $1.50 or $1 and 50cents to play one game

Step-by-step explanation:

let x represent cost of games and y represent cost of rides

6x+4y=$22——equation 1

4x+8y=$32——equation 2

Multiply equation 1by 2

2(6x+4y=$22)=12x+8y=$44

Subtract equation 2from 1 to eliminate y

12x-4x+8y-8y=$44-32

8x=$12

x=$12/8=$1.5

Plug in x=1.5 into equation 2

4($1.5)+8y=$32

8y=$32-$6

8y=$26

y=26/8=$3.25

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What are the measures of the missing angles x°, y° of a regular triangle if one angle measure equals 60°?​
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Answer:

So each interior angle measurement of the regular triangle is 60 degrees.

Step-by-step explanation:

If the triangle is regular, this means all the side measurements are congruent to each other and that all the angle measurements are congruent to each other.

If the sum of the angles in a triangle is 180 degrees and you know they are each the same then you could either solve x+x+x=180 or know we are just dealings with an equiangular triangle in which all angles have measurement 60 degrees.

If need more convincing, let's actually solve:

x+x+x=180

3x=180

x=180/3

x=60

So each interior angle measurement of the regular triangle is 60 degrees.

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Hope this helps :)
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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}

\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

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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

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