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mote1985 [20]
3 years ago
5

Explain how can you determine if a solid has reflectional symmetry.

Mathematics
1 answer:
Pani-rosa [81]3 years ago
4 0

Answer: When a solid can be mapped onto itself by a reflection in a plane, it has reflectional symmetry.

Step-by-step explanation:

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A formula for finding SA, the surface area of a rectangular prism, is
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Answer:

first you need to make your shape to always get you stuff right and that is why the d is the right one.

Step-by-step explanation:

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a cog company produces 20 cogs a day, 4 of which are defective. Find the probability of selecting 4 cogs from the 20 produced wh
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Answer:

1/4845

Step-by-step explanation:

The probability of selecting a defective cog first is

1/5

The second cog is

3/19

The third cog is

1/9

And the fourth cog is

1/17

Multiplying these together we get

1/5 * 3/19* 1/9 * 1/17  = 1/4845

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3 years ago
I need to know the answer
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Slope = rise over run = the change in y over the change of x
if you want it to be exactly 2, then you have to make sure the equations are proportional.
 -5 - y / 6-1
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8 0
3 years ago
A mass weighing 16 pounds stretches a spring (8/3) feet. The mass is initially released from rest from a point 2 feet below the
mezya [45]

Answer with Step-by-step explanation:

Let a mass weighing 16 pounds stretches a spring \frac{8}{3} feet.

Mass=m=\frac{W}{g}

Mass=m=\frac{16}{32}

g=32 ft/s^2

Mass,m=\frac{1}{2} Slug

By hook's law

w=kx

16=\frac{8}{3} k

k=\frac{16\times 3}{8}=6 lb/ft

f(t)=10cos(3t)

A damping force is numerically equal to 1/2 the instantaneous velocity

\beta=\frac{1}{2}

Equation of motion :

m\frac{d^2x}{dt^2}=-kx-\beta \frac{dx}{dt}+f(t)

Using this equation

\frac{1}{2}\frac{d^2x}{dt^2}=-6x-\frac{1}{2}\frac{dx}{dt}+10cos(3t)

\frac{1}{2}\frac{d^2x}{dt^2}+\frac{1}{2}\frac{dx}{dt}+6x=10cos(3t)

\frac{d^2x}{dt^2}+\frac{dx}{dt}+12x=20cos(3t)

Auxillary equation

m^2+m+12=0

m=\frac{-1\pm\sqrt{1-4(1)(12)}}{2}

m=\frac{-1\pmi\sqrt{47}}{2}

m_1=\frac{-1+i\sqrt{47}}{2}

m_2=\frac{-1-i\sqrt{47}}{2}

Complementary function

e^{\frac{-t}{2}}(c_1cos\frac{\sqrt{47}}{2}+c_2sin\frac{\sqrt{47}}{2})

To find the particular solution using undetermined coefficient method

x_p(t)=Acos(3t)+Bsin(3t)

x'_p(t)=-3Asin(3t)+3Bcos(3t)

x''_p(t)=-9Acos(3t)-9sin(3t)

This solution satisfied the equation therefore, substitute the values in the differential equation

-9Acos(3t)-9Bsin(3t)-3Asin(3t)+3Bcos(3t)+12(Acos(3t)+Bsin(3t))=20cos(3t)

(3B+3A)cos(3t)+(3B-3A)sin(3t)=20cso(3t)

Comparing on both sides

3B+3A=20

3B-3A=0

Adding both equation then, we get

6B=20

B=\frac{20}{6}=\frac{10}{3}

Substitute the value of B in any equation

3A+10=20

3A=20-10=10

A=\frac{10}{3}

Particular solution, x_p(t)=\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)

Now, the general solution

x(t)=e^{-\frac{t}{2}}(c_1cos(\frac{\sqrt{47}t}{2})+c_2sin(\frac{\sqrt{47}t}{2})+\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)

From initial condition

x(0)=2 ft

x'(0)=0

Substitute the values t=0 and x(0)=2

2=c_1+\frac{10}{3}

2-\frac{10}{3}=c_1

c_1=\frac{-4}{3}

x'(t)=-\frac{1}{2}e^{-\frac{t}{2}}(c_1cos(\frac{\sqrt{47}t}{2})+c_2sin(\frac{\sqrt{47}t}{2})+e^{-\frac{t}{2}}(-c_1\frac{\sqrt{47}}{2}sin(\frac{\sqrt{47}t}{2})+\frac{\sqrt{47}}{2}c_2cos(\frac{\sqrt{47}t}{2})-10sin(3t)+10cos(3t)

Substitute x'(0)=0

0=-\frac{1}{2}\times c_1+10+\frac{\sqrt{47}}{2}c_2

\frac{\sqrt{47}}{2}c_2-\frac{1}{2}\times \frac{-4}{3}+10=0

\frac{\sqrt{47}}{2}c_2=-\frac{2}{3}-10=-\frac{32}{3}

c_2==-\frac{64}{3\sqrt{47}}

Substitute the values then we get

x(t)=e^{-\frac{t}{2}}(-\frac{4}{3}cos(\frac{\sqrt{47}t}{2})-\frac{64}{3\sqrt{47}}sin(\frac{\sqrt{47}t}{2})+\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)

8 0
3 years ago
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