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3 years ago

For what numbers theta is f(theta)=cot(theta) not defined?

1 answer:

5 0

as cosec theeta is inverse of sin theeta:
so
csc0=1/sin0
and sin0=0 when 0={0.pi,and 2pi}
when 0=0.pi,and 2pi

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Three farmers buy a total of 2,249 acres of land at an auction, Farmer Mel buys land that is 3 times as many acres as Farmer Cas

fgiga [73]

Answer:

Farmer Anne's land = 173 acres

Farmer Cassius = 519 acres

Farmer Mel = 1,557 acres

Step-by-step explanation:

Total acre = 2,249

Let

Farmer Anne's land = x

Farmer Cassius = 3x

Farmer Mel = 3(3x) = 9x

x + 3x + 9x = 2,249

13x = 2,249

x = 173

Farmer Anne's land = x = 173 acres

Farmer Cassius = 3x

= 3(173)

= 519 acres

Farmer Mel = 3(3x) = 9x

= 9(173)

= 1,557 acres

Therefore,

Farmer Anne's land = 173 acres

Farmer Cassius = 519 acres

Farmer Mel = 1,557 acres

8 0

2 years ago

A gas is said to be compressed adiabatically if there is no gain or loss of heat. When such a gas is diatomic (has two atoms per

Tems11 [23]

Answer:

The pressure is changing at $\frac{dP}{dt}=3.68$

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of $\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}$ and we want to find at what rate is the pressure changing.

The equation that model this situation is

$PV^{1.4}=k$

Differentiate both sides with respect to time t.

$\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\$

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

$\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)$

Apply this rule to our expression we get

$V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0$

Solve for $\frac{dP}{dt}$

$V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}$