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Marrrta [24]
3 years ago
6

Place the following steps in order to describe how to solve the following equation: ​ ​ 2(2x+3)-2=5

Mathematics
1 answer:
ivolga24 [154]3 years ago
5 0

Answer:

0.25

Step-by-step explanation:

First, you distribute the equation.

2(2x+3)-2=5

4x+6-2=5

Then, you combine like terms.

4x+6-2=5

4x+4=5

Next, you subtract 4 from each side.

4x+4=5

4x+4-4=4x

5-4=1

4x=1

Finally, you divide each side by 4.

4x/4

1/4

(4x/4 you'd end up with x)

Answer: x=1/4

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Let her sisters age be x.
=>sister .....x

Sarah.
twice x=2x. then less than 5
=> (2x-5 )
if Sarah = 15 years
by substation, 2x-5=15
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4 years ago
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Assume each figure shown has the same orientation. Which figure is the image of square LMNP after a translation of
umka21 [38]

Figure 4 is the image of the square LMNP after the translation.

<u>Step-by-step explanation:</u>

Let us see the coordinates of the pre image LMNP as,

L (-3,1)

M(-1,1)

N(-1,-1)

P(-3,-1)

after translation of (x,y) → (x+5, y -3) the coordinates of the image obtained as,

L'(2,-2)

M'(4,-2)

N'(4,-4)

P'(2,-4) which matches the image 4.

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3 years ago
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A ball slides up a frictionless ramp. It is then rolled without slipping and with the same initial velocity up another frictionl
KATRIN_1 [288]

Answer:

Rolling case achieves greater height than sliding case

Step-by-step explanation:

For sliding ball:

- When balls slides up the ramp the kinetic energy is converted to gravitational potential energy.

- We have frictionless ramp, hence no loss due to friction.So the entire kinetic energy is converted into potential energy.

- The ball slides it only has translational kinetic energy as follows:

                                   ΔK.E = ΔP.E

                                   0.5*m*v^2 = m*g*h

                                    h = 0.5v^2 / g

For rolling ball:

- Its the same as the previous case but only difference is that there are two forms of kinetic energy translational and rotational. Thus the energy balance is:

                                  ΔK.E = ΔP.E

                                  0.5*m*v^2 + 0.5*I*w^2 = m*g*h

- Where I: moment of inertia of spherical ball = 2/5 *m*r^2

             w: Angular speed = v / r

                               0.5*m*v^2 + 0.2*m*v^2 = m*g*h

                               0.7v^2 = g*h

                               h = 0.7v^2 / g

- From both results we see that 0.7v^2/g for rolling case is greater than 0.5v^2/g sliding case.

4 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

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