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ludmilkaskok [199]
3 years ago
11

What is the following sum? 2 (RootIndex 3 StartRoot 16 x cubed y EndRoot) 4 (RootIndex 3 StartRoot 54 x Superscript 6 Baseline y

Superscript 5 Baseline) 4 x (RootIndex 3 StartRoot 2 y EndRoot) 12 x squared y (RootIndex 3 StartRoot 2 y squared EndRoot) 8 x (RootIndex 3 StartRoot x y EndRoot) 12 x cubed y squared (RootIndex 3 StartRoot 6 y EndRoot) 16 x cubed y (RootIndex 3 StartRoot 2 y squared EndRoot) 48 x cubed y (RootIndex 3 StartRoot 2 y EndRoot).
Mathematics
2 answers:
san4es73 [151]3 years ago
7 0

Equivalent expressions are expressions that have the same value, and can be used interchangeably.

The result of the sum 2 (\sqrt[3]{16x^3y})  + 4 (\sqrt[3]{54x^6y^5}) is 4x\sqrt[3]{2y}  + 8x^2y\sqrt[3]{2y^2})

The expression is given as:

2 (\sqrt[3]{16x^3y})  + 4 (\sqrt[3]{54x^6y^5})

Rewrite the expression as:

2 (\sqrt[3]{16x^3y})  + 4 (\sqrt[3]{54x^6y^5}) = 2 (\sqrt[3]{2^4x^3y})  + 4 (\sqrt[3]{3^3 \times 2x^6y^5})

Evaluate the roots

2 (\sqrt[3]{16x^3y})  + 4 (\sqrt[3]{54x^6y^5}) = 2 (2x\sqrt[3]{2y})  + 4 (3x^2y\sqrt[3]{2y^2})

Open the brackets

2 (\sqrt[3]{16x^3y})  + 4 (\sqrt[3]{54x^6y^5}) = 4x\sqrt[3]{2y}  + 12x^2y\sqrt[3]{2y^2})

The above expression cannot be further simplified.

Hence, the result of the sum 2 (\sqrt[3]{16x^3y})  + 4 (\sqrt[3]{54x^6y^5}) is 4x\sqrt[3]{2y}  + 8x^2y\sqrt[3]{2y^2})

Read more about equivalent expressions at:

brainly.com/question/2972832

yulyashka [42]3 years ago
4 0

The sum of the expression is 4 (\sqrt[3]{x^3y} +12x^2y(\sqrt[3]{ 2 y^2})\\.

We have to determine, the sum of the given expression.

According to the question,

Expression; 2(\sqrt[3]{16x^3y} +4(\sqrt[3]{56x^6y^5})

To determine the sum of the given expression following all the steps given below.

Rewrite the expression term in the form of their cubes,

\rm = 2(\sqrt[3]{16x^3y} +4(\sqrt[3]{56x^6y^5})\\\\ = 2(\sqrt[3]{2^4x^3y} +4(\sqrt[3]{3^3 \times 2 \times x^6y^5})\\\\= 2\times 2(\sqrt[3]{x^3y} +4\times 3(\sqrt[3]{ 2 \times x^6y^5})\\\\= 4 (\sqrt[3]{x^3y} +12x^2y(\sqrt[3]{ 2 y^2})\\\\=4 (\sqrt[3]{x^3y} +12x^2y(\sqrt[3]{ 2 y^2})\\

Hence, The sum of the expression is 4 (\sqrt[3]{x^3y} +12x^2y(\sqrt[3]{ 2 y^2})\\.

For more details refer to the link given below.

brainly.com/question/21798224

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V = (1/3)(s²)(2/3)s, or V = (2/9)s³

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A mass of 1 g is set in motion from its equilibrium position with an initial velocity of 6in/sec, with no damping and a spring c
yan [13]

a) y(t)=0.0016 sin(94.9t) [m]

b) 0.033 s

c) -0.152 m/s

Step-by-step explanation:

a)

The force acting on the mass-spring system is (restoring force)

F=-ky

where

k = 9 is the spring constant

y is the displacement

Also, from Newton's second law of motion, we know that

F=my''

where

m = 1 g = 0.001 kg is the mass

y'' is the acceleration

Combining the two equations,

my''=-ky

This is a second order differential equation; the solution for y(t) is

y(t)=A sin(\omega t-\phi)

where

A is the amplitude of motion

\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{9}{0.001}}=94.9 rad/s is the angular frequency

The spring starts its motion from its equilibrium position, this means that y=0 when t=0; therefore, the phase shift must be

\phi=0

So the displacement is

y(t)=A sin(\omega t)

The velocity of the spring is equal to the derivative of the displacement:

v(t)=y'(t)=\omega A cos(\omega t)

We know that at t = 0, the initial velocity is 6 in/s; since 1 in = 2.54 cm = 0.0254 m,

v_0=6(0.0254)=0.152 m/s

And since at t = 0, cos(\omega t)=1

Then we have:

v_0=\omega A

From which we find the amplitude:

A=\frac{v_0}{\omega}=\frac{0.152}{94.9}=0.0016 m

So the solution for the displacement is

y(t)=0.0016 sin(94.9t) [m]

b)

Here we want to find the time t at which the mass returns to equilibrium, so the time t at which

y=0

This means that

sin(\omega t)=0

We know already that the first time at which this occurs is

t = 0

Which is the beginning of the motion.

The next occurence of y = 0 is instead when

\omega t = \pi

which means:

t=\frac{\pi}{\omega}=\frac{\pi}{94.9}=0.033 s

c)

As said in part a), the velocity of the mass-spring system at time t is given by the derivative of the displacement, so

v(t)=\omega A cos(\omega t)

where we have

\omega=94.9 rad/s is the angular frequency

A=0.0016 m is the amplitude of motion

t is the time

Here we want to find the velocity of the mass when the time is that calculated in part b):

t = 0.033 s

Substituting into the equation, we find:

v(0.033)=(94.9)(0.0016)cos(94.9\cdot 0.033)=-0.152 m/s

4 0
3 years ago
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