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2 years ago

Use the figure to find n please.

Mathematics

2 answers:

ira [324]2 years ago

4 0

Answer:

n = 5

Step-by-step explanation:

Since this is a right triangle, we can use trig

tan theta = opp /adj

tan 30 = n / 5 sqrt(3)

5 sqrt(3) tan 30 = n

5 sqrt(3) * 1/ sqrt(3) = n

5 = n

Sergio039 [100]2 years ago

3 0

Now we have to,

find the required value of n.

Now we can,

Use the trigonometric functions.

→ tan(θ) = opp/adj

Let's find the required value of n,

→ tan (θ) = opp/adj

→ tan (30) = n/5√3

→ n = 5√3 × tan (30)

→ n = 5√3 × √3/3

→ n = 5√3 × 1/√3

→ [n = 5]

Hence, the value of n is 5.

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Step-by-step explanation:

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2 years ago

Find the solution of the equation. 2(2x + 3) = -4 + 2(2x + 5)

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3 years ago

Read 2 more answers

A tank contains 240 liters of fluid in which 20 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu

Novosadov [1.4K]

Answer:

$A(t)=240-220e^{-\frac{t}{40}}$

Step-by-step explanation:

A tank contains 240 liters of fluid in which 20 grams of salt is dissolved.

Volume of the tank = 240 liters
Initial Amount of Salt in the tank, A(0)=20 grams

Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 6 L/min

$R_{in}=$ (concentration of salt in inflow)(input rate of fluid)

$R_{in}=(1\frac{gram}{liter})( 6\frac{Liter}{min})=6\frac{gram}{min}$

$R_{out}=$ (concentration of salt in outflow)(output rate of fluid)

$R_{out}=(\frac{A(t)}{240})( 6\frac{Liter}{min})\\R_{out}=\frac{A}{40}$

Rate of change of the amount of salt in the tank:

$\dfrac{dA}{dt}=R_{in}-R_{out}$

$\dfrac{dA}{dt}=6-\dfrac{A}{40}$

We then solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{40}=6\\$The integrating factor: e^{\int \frac{1}{40}dt} =e^{\frac{t}{40}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{40}}+\dfrac{A}{40}e^{\frac{t}{40}}=6e^{\frac{t}{40}}\\(Ae^{\frac{t}{40}})'=6e^{\frac{t}{40}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{40}})'=\int 6e^{\frac{t}{40}} dt\\Ae^{\frac{t}{40}}=6*40e^{\frac{t}{40}}+C, $(C a constant of integration)\\Ae^{\frac{t}{40}}=240e^{\frac{t}{40}}+C\\$Divide all through by e^{\frac{t}{40}}\\A(t)=240+Ce^{-\frac{t}{40}}$

Recall that when t=0, A(t)=20 (our initial condition)

$20=240+Ce^{-\frac{0}{40}}\\20-240=C\\C=-220\\$Therefore, the number A(t) of grams of salt in the tank at time t\\A(t)=240-220e^{-\frac{t}{40}}$

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3 years ago

The distance between two cities on a map is 3 1/2 inches. The actual distance between the two cities is 28 miles.

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Part A= 3.5 inches is equal to 8 miles

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3 years ago

The price of a technology stock was $9.74 yesterday. Today, the price fell to $9.67. Find the percentage decrease. Round your an

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3 years ago