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3 years ago

Find the value of y round your answer to the nearest tenth cos y y=8/13

Mathematics

2 answers:

Iteru [2.4K]3 years ago

7 0

8/13 = 0.6153

y = cos-1(0.6153)

y = 52.0

I'm not totally sure if this is what you mean or if this is.

y = cos(8/13) = 0.9999 which to the nearest tenth is 1.0

weqwewe [10]3 years ago

3 0

Answer:

52.0

Step-by-step explanation:

To solve this, we will perform the inverse cosine:

cos y = 8/13

cos⁻¹(cos y) = cos⁻¹(8/13)

y = 52.0

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2 years ago

A car dealership paid $12,800 for a car. They mark up the price by 20%

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2 years ago

Read 2 more answers

A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and

goldfiish [28.3K]

Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

$\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}$

Solve for S(t):

$\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}$

$e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}$

The left side is the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}$

Integrate both sides:

$e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt$

$e^{t/800}S(t)=64e^{t/800}+C$

$S(t)=64+Ce^{-t/800}$

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

$0=64+C\impleis C=-64$

and so (b) the amount of sugar in the tank at time t is

$S(t)=64\left(1-e^{-t/800}\right)$

As $t\to\infty$ , the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

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3 years ago