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Sidana [21]
3 years ago
11

What term describes the "3" in the expression 3 × 2 ÷ 1?

Mathematics
1 answer:
Stolb23 [73]3 years ago
5 0

Answer:

3 represents by poloar experience.

Step-by-step explanation:

by polar is a convicted number with a 4 by 4 grid conjuncted with a 7 by 9 terform grapple 20 by 20 grip.

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Suppose you have developed a scale that indicates the brightness of sunlight . Each category in the tablet is 5 time brighter th
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If you could put in the tablet.

But the equation would be: a_n = a_1 (r)^(x-1)

with a_1 being the first category

r being 5 how much brighter than the day before is

and x is what nr. in the sequence it is

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8 0
2 years ago
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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
Expand and simplify 5(y-2) 2(y-3)
Oksana_A [137]
5y-10 and 2y-6 unless the two are together in which case are they being added or multiplied (the equations).

4 0
3 years ago
The ordered pair (x – 3, y + 1) was used to translate an object. To find the pre-image of the object, you would need to move ___
lisov135 [29]
B. three units down and one unit right

x minus will always translate down, x plus will always translate up.

y minus will always translate left, y plus will always translate right.
8 0
2 years ago
5-2.1=? 5-2.17=? 5-2 7/8 solve make sure to re-check your self before answering or no forty points
frozen [14]

Answer: 5-2.1= 2.9

5-2.17= 2.83

5-2 7/8= 2.125

6 0
3 years ago
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