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uranmaximum [27]
2 years ago
8

Between the numbers 15/20 and 35/40 , the greater number is a. 15/20 b. 20/15 c. 35/40 d. 45/30

Mathematics
2 answers:
Citrus2011 [14]2 years ago
8 0

Answer:

35/40

Step-by-step explanation:

To compare, both denominator should be same.

\frac{15}{20}=\frac{15*2}{20*2}=\frac{30}{40}\\\\\\\frac{30}{40} \ < \ \frac{35}{40}\\\\\\\frac{15}{20} \ < \ \frac{35}{40}

HACTEHA [7]2 years ago
8 0

\sf{\bold{\blue{\underline{\underline{Given}}}}}

  • ⠀the numbers 15/20 and 35/40⠀⠀⠀

\sf{\bold{\red{\underline{\underline{To\:Find}}}}}

⠀⠀⠀⠀

  • the greater number is

\sf{\bold{\purple{\underline{\underline{Solution}}}}}

we have to find the greater number between 15/20 and 35/40

To do this,

we have to compare the denominator same.

  • \sf{\dfrac{35}{40}=\dfrac{35×1}{40×1}=\dfrac{35}{40}   }

  • \sf{\dfrac{15}{20}=\dfrac{15×2}{20×2}=\dfrac{30}{40}   }

<u>According</u><u> </u><u>to</u><u> </u><u>the question</u><u>, </u>

we have to find the greatest one

  • \sf{\dfrac{35}{40} > \dfrac{30}{40}  }

  • \sf{\dfrac{35}{40}>\dfrac{15}{20}   }

\sf{\bold{\green{\underline{\underline{Answer}}}}}

⠀⠀

\therefore\mathrm{\dfrac{35}{40}  >  \dfrac{15}{20} }

⠀⠀

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lord [1]
Let N be no. of nickles. Then N($.05)+3N($.10)+2N($.25)=$3.40.

Or

4 nickels = $0.20
12 dimes = $1.20
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Total = $3.40

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What is the value of X?
nasty-shy [4]

Answer:

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7 0
3 years ago
A mass of 3.25 kg is attached to the end of a spring that is stretched 22 cm by a force of 15 N. It is set in motion with an ini
mylen [45]

Answer:

Step-by-step explanation:

Given that,

Mass of object=3.25kg

The extension e=22cm=0.22m

Force applied to cause extension F=15N

Initial position Xo=0

Initial velocity Vo=-12m/s

We can get the spring constant from Hooke's law

F=ke

Then, k=F/e

k=15/0.22

k=68.182N/m

Also our natural frequency w is given as

w=√(k/m)

Therefore,

w=√(68.182/3.24)

w=√20.98

w=√21

w=4.58rad/s

w=4.6rad/s

There is no damping in this situation, no outside force acting on the system and the equation that governs the system is

mx''+kx=0

3.25x''+68.182x=0

Divide through by 3.25

x''+20.98x=0

We can approximate 20.98 to 21

x"+21x=0

The solution to this differential equation using D operator

D²+21=0

D²=-21

D= ±√-21

D=±√21 •i

Then the solution is

x(t)=A•Sinwt +B•Coswt

x(t)=A•Sin√21 t +B•Cos√21 t

Note that x'(t)=v(t)

and at t=0 Vo=-12m/s

x(t)=A•Sin√21 t +B•Cos√21 t

x'(t)=v(t)=A√21•Cos√21 t - B√21•Sin√21 t

v(t)=A√21•Cos√21 t - B√21•Sin√21 t

Then, using the two initial conditions

v(0)=-12

And X(0)=0

x(t)=A•Sin√21 t +B•Cos√21 t

X(0)=A•Sin√21•0 +B•Cos√21•0

X(0)=A•Sin0+B•Cos0

0=B

B=0

Also,, V(0)=-12m/s

v(t)=A√21•Cos√21 t - B√21•Sin√21 t

V(0)=A√21•Cos√21•0- B√21•Sin√21•0

V(0)=A√21•Cos0- B√21•Sin0

-12=A√21

Therefore,

A=-12/√21

A=-2.62

Therefore the general equation becomes

x(t)=A•Sin√21 t +B•Cos√21 t

x(t)=-2.62Sin√21 t +0•Cos√21 t

x(t)=-2.62Sin√21 t

a. The amplitude

Comparing x(t) to wave equations

x(t)=-Asin(wt+2λ/t)

Then,

A=2.62m

b. We know the natural frequency already to be

w=√21

w=4.58rad/s

c. Period

Comparing the equation again

wt=√21t

Given that w=2πf

Therefore, 2πft=√21t

Then, f=√21t / 2πt

f=√21/2π

f=0.73Hz

Then, period is the reciprocal of frequency

T=1/f

T=1/0.73

T=1.37seconds

The period is 1.37sec,

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