What does it mean to empathize with a monster?A. To ignore the monster's motivesB. To feel what the monster is feelingC. To suspect the monster's wordsD. To hate the monster rather than fear it​

Answer:

Control the strategy of the party

Explanation:

When you are the leader of a party in the legislative branch, you have the power to control the strategy of the party.

Roles of Party Leaders

• Manage the strategy of a party.
• Act as spokespeople for the party.
• Manage party relationship with the electorate.

As the ones who manage/ control the strategy of the party, party leaders have immense influence when it comes to deciding how party members vote.

In conclusion, option B is correct.

Answer:

D

Step-by-step explanation:

x/(x²+3x+2) - 1/[(x+2)(x+1)]

x² + 3x + 2 = x² + 2x + x + 2

= x(x + 2) + (x + 2)

= (x + 2)(x + 1)

= x/[(x+2)(x+1)] - 1/[(x+2)(x+1)]

= (x-1)/[(x+2)(x+1)]

= (x-1)/(x²+3x+2)

The difference of rational polynomial function is \frac{x-1}{x^{2} +3x+2}

Given expression is,

          \frac{x}{x^{2} +3x+2}-\frac{1}{(x+2)(x+1)}

(x+2)(x+1)=x^{2} +2x+x+2=x^{2} +3x+2

Given expression can be written as,

       \frac{x}{x^{2} +3x+2}-\frac{1}{x^{2} +3x+2} =\frac{x-1}{x^{2} +3x+2}

Answer:

The time rate of change in air density during expiration is 0.01003kg/m³-s

Explanation:

Given that,

Lung total capacity V = 6000mL = 6 × 10⁻³m³

Air density p = 1.225kg/m³

diameter of the trachea is 18mm = 0.018m

Velocity v = 20cm/s = 0.20m/s

dv /dt = -100mL/s (volume rate decrease)

= 10⁻⁴m³/s

Area for trachea =

\frac{\pi }{4} d^2\\= 0.785\times 0.018^2\\= 2.5434 \times10^-^4m^2

0 - p × Area for trachea =

\frac{d}{dt} (pv)=v\frac{ds}{dt} + p\frac{dv}{dt}

-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)

-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)

⇒-0.623133\times10^-^4+1.225\times10^-^4=6\times10^-^3\frac{ds}{dt}

           \frac{ds}{dt} = \frac{0.6018\times10^-^4}{6\times10^-^3} \\\\= 0.01003kg/m^3-s

ds/dt = 0.01003kg/m³-s

Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s

Answer:

Time rate = 0.010026 kg/m³.s

Explanation:

We are given the following;

Total lung capacity; V = 6000mL = 6 x 10^(-3) m³

diameter of the trachea is 18mm = 0.018m

Air density; ρ = 1.225kg/m³

Velocity; v = 20cm/s = 0.20m/s

From the question, lung volume is decreasing at a rate of 100mL/s.

Thus, dv /dt = -100mL = -0.0001 m³/s

Area of trachea; A = πD²/4

A = (π x 0.018²)/4 = 2.5447 x 10^(-4) m²

Now, let's set up the equation.

-ρAv = (d/dt)(ρV) = V(dρ/dt) + ρ(dv/dt)

Thus,

Plugging in the relevant values to get;

-[1.225 x 2.5447 x 10^(-4) x 0.20] = 6 x 10^(-3)(dρ/dt) + (1.225 x -0.0001)

So,

-0.62345 x 10^(-4) = 6x10^(-3)(dρ/dt) - (1.225 x 10^(-4))

6x10^(-3)(dρ/dt) = (1.225 x 10^(-4)) - 0.62345 x 10^(-4)

6x10^(-3)(dρ/dt) = 0.60155 x 10^(-4)

(dρ/dt) = [0.60155 x 10^(-4)] /(6x10^(-3)) = 0.10026 x 10^(-1) = 0.010026 kg/m³.s

ρ