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Lostsunrise [7]
2 years ago
12

You can use a system of equations to graph and solve the polynomial equation 3 x cubed + x = 2 x squared + 1. Which statement is

true? The system has one solution, and the equation has three zeroes. The y-coordinates of the solutions to the system and the zeroes of the equation are not equal. The x-coordinates of the solutions to the system and the zeroes of the equation are not equal. The equation has one zero, and the system has three solutions.
Physics
1 answer:
GuDViN [60]2 years ago
5 0

The equation has one zero, and the system has three solutions ( 1 real and 2 complex solutions).

<h3>Solution of the polynomial equation</h3>

The solution of the polynomial equation is determined as follows;

3x³ + x = 2x² + 1

3x³ - 2x² + x - 1 = 0

3(x - \frac{2}{9} )^3+ \frac{5}{9} (x - \frac{2}{9} )- \frac{205}{243} = 0\\\\x = 0.78 \ (one  \ real \ solution)\\\\x = -0.058-0.65i\\\\x = -0.058+0.65i \ \ (2  \ complex \ solutions)

Thus, we can conclude that the equation has one zero (real solution), and the system has three solutions ( 1 real and 2 complex solutions).

Learn more about polynomial equations here: brainly.com/question/2833285

#SPJ1

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Now explore friction force. Set the piece of plastic or wood on the table and push it steadily across the tabletop using your fi
ehidna [41]

Answer:

The tabletop is smooth so my finger is down it fast and easy. The fabric however slowed my finger down considerably, and it was harder for me to move my finger across it.

Explanation:

Hope this helps.

3 0
2 years ago
the density of aluminum is 2700 kg/m3. if transverse waves travel at in an aluminum wire of diameter what is the tension on the
Sunny_sXe [5.5K]

The tension on the wire is 52.02 N.

From the question, we have

Density of aluminum = 2700 kg/m3

Area,

A = πd²/4

A = π x (4.6 x 10⁻³)²/4

A = 1.66 x 10⁻⁵ m²

μ = Mass per unit length of the wire

μ = ρA

μ = 2700 kg/m³ x 1.66 x 10⁻⁵ m²

μ = 0.045 kg/m

Tension on the wire = √T/μ

34 = √T/0.045

34² = T/0.045

T = 52.02 N

The tension on the wire is 52.02 N.

Complete question:

The density of aluminum is 2700 kg/m3. If transverse waves propagate at 34 m/s in a 4.6-mm diameter aluminum wire, what is the tension on the wire.

To learn more about tension visit:  brainly.com/question/14336853

#SPJ4

5 0
1 year ago
A device used to measure the atmospheric pressure is the barometer. true false
Angelina_Jolie [31]
True, I just learned this a week ago. Is this for Chemistry?
3 0
3 years ago
(a) An elevator of mass m moving upward has two forces acting on it: the upward force of tension in the cable and the downward f
Katen [24]

Answer: T is greater

Explanation:

Since the elevator is moving against gravity more work will be done on the rope

T= m(g+a)

8 0
3 years ago
Read 2 more answers
3. A 2kg wooden block whose initial speed is 3 m/s slides on a smooth floor for 2 meters before it comes to a
serious [3.7K]

Answer:

Calculating Coefficient of friction is 0.229.

Force is 4.5 N that keep the block moving at a constant speed.

Explanation:

We know that speed expression is as \mathrm{V}^{2}=\mathrm{V}_{\mathrm{i}}^{2}+2 . \mathrm{a} . \Delta \mathrm{s}.

Where, {V}_{i} is initial speed, V is final speed, ∆s displacement and a acceleration.

Given that,

{V}_{i} =3 m/s, V = 0 m/s, and  ∆s = 2 m

Substitute the values in the above formula,

0=3^{2}-2 \times 2 \times a

0 = 9 - 4a

4a = 9

a=2.25 \mathrm{m} / \mathrm{s}^{2}

a=2.25 \mathrm{m} / \mathrm{s}^{2} is the acceleration.

Calculating Coefficient of friction:

\mathrm{F}=\mathrm{m} \times \mathrm{a}

\mathrm{F}=\mu \times \mathrm{m} \times \mathrm{g}

Compare the above equation

\mu \times m \times g=m \times a

Cancel "m" common term in both L.H.S and R.H.S

\text { Equation becomes, } \mu \times g=a

\text { Coefficient of friction } \mu=\frac{a}{g}

\mathrm{g} \text { on earth surface }=9.8 \mathrm{m} / \mathrm{s}^{2}

\mu=\frac{2.25}{9.8}

\mu=0.229

Hence coefficient of friction is 0.229.

calculating force:

\text { We know that } \mathrm{F}=\mathrm{m} \times \mathrm{a}

\mathrm{F}=2 \times 2.25 \quad(\mathrm{m}=2 \mathrm{kg} \text { given })

F = 4.5 N

Therefore, the force would be <u>4.5 N</u> to keep the block moving at a constant speed across the floor.

7 0
3 years ago
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