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2 years ago

Obtain the zeroes of polynomial

Physics

1 answer:

morpeh [17]2 years ago

3 0

I assume you meant to say

$f(x)=2x^4+3x^3-5x^2-9x-3$

Given that x = √3 and x = -√3 are roots of f(x), this means that both x - √3 and x + √3, and hence their product x ² - 3, divides f(x) exactly and leaves no remainder.

Carry out the division:

$\dfrac{2x^4+3x^3-5x^2-9x-3}{x^2-3} = 2x^2+3x+1$

To compute the quotient:

* 2x ⁴ = 2x ² • x ², and 2x ² (x ² - 3) = 2x ⁴ - 6x ²

Subtract this from the numerator to get a first remainder of

(2x ⁴ + 3x ³ - 5x ² - 9x - 3) - (2x ⁴ - 6x ²) = 3x ³ + x ² - 9x - 3

* 3x ³ = 3x • x ², and 3x (x ² - 3) = 3x ³ - 9x

Subtract this from the remainder to get a new remainder of

(3x ³ + x ² - 9x - 3) - (3x ³ - 9x) = x ² - 3

This last remainder is exactly divisible by x ² - 3, so we're left with 1. Putting everything together gives us the quotient,

2x ² + 3x + 1

Factoring this result is easy:

2x ² + 3x + 1 = (2x + 1) (x + 1)

which has roots at x = -1/2 and x = -1, and these re the remaining zeroes of f(x).

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A jet accelerates at a=3.92 m/s2 from rest until it reaches its takeoff velocity of 360 km/hr. It has to travel for 5 seconds at

Art [367]

Answer:

$\Delta s = 1775.510\,m$

Explanation:

The minimum distance for takeoff is:

$\Delta s = \Delta s_{1} + \Delta s_{2}$

$\Delta s = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot a} + v_{f}\cdot \Delta t$

$\Delta s = \frac{(100\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}}{2\cdot (3.92\,\frac{m}{s^{2}})}+(100\,\frac{m}{s} )\cdot (5\,s)$

$\Delta s = 1775.510\,m$

7 0

2 years ago

For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par

Vitek1552 [10]

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

initial temperature of metals, = $T_m$
initial temperature of water, = $T_i$
specific heat capacity of copper, $C_p$ = 0.385 J/g.K
specific heat capacity of aluminum, $C_A$ = 0.9 J/g.K
both metals have equal mass = m

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

For copper metal, the quantity of heat transferred is calculated as;

$Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w ) = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t$

The change in heat energy for copper metal;

$\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t$

For aluminum metal, the quantity of heat transferred is calculated as;

$Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w) = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t$

The change in heat energy for aluminum metal ;

$\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t$

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

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6 0

2 years ago

What voltage produces a current of 50 amps with a resistance of 20 ohms?

PIT_PIT [208]

Answer:

1000 V

Explanation:

Ohm's law:

V = IR

V = (50 A) (20 Ω)

V = 1000 V

5 0

3 years ago

I'm stuck on number 4... help please?

Paha777 [63]

Average speed = (distance covered) / (time to cover the distance) =

 (5m) / (15 sec) =

 (5/15) (m/s) = 1/3 m/s .

Average velocity =

 (displacement) / (time spent traveling) in the direction of the displacement

Average velocity = (5m) / (15 sec) left =

 (5/15) / (m/sec) left =

 1/3 m/s left.

A number without a direction is a speed, not a velocity.

3 0

3 years ago

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A thermometer containing 0.10 g of mercury is cooled from 15 degrees celsius to 8.5 degrees celcius. How much energy left the me

loris [4]

To solve this exercise we will use the concept related to heat loss which is mathematically given as

$Q = mC_p \Delta T$