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2 years ago

Question 32 of 40 The ratio opposite hypotenuse describes which function?

Mathematics

1 answer:

Helen [10]2 years ago

5 0

The ratio $\frac{Opposite}{Hypotenuse }$ describes the Sine function . The correct option is C. Sine

<h3>Trigonometry </h3>

From the question, we are to determine the function that the ratio of <em>opposite </em>to <em>hypotenuse </em>describes

By SOH CAH TOA

We have that

$Sine = \frac{Opposite}{Hypotenuse }$

$Cosine = \frac{Adjacent}{Hypotenuse }$

and

$Tangent = \frac{Opposite}{Adjacent}$

Hence, the ratio $\frac{Opposite}{Hypotenuse }$ describes the Sine function . The correct option is C. Sine

Learn more on Trigonometry here: brainly.com/question/956353

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3 years ago

Table of values 3x+2y=24

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−
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2
-
3
2
Y-Intercept:
12
12
Graph a dashed line, then shade the area above the boundary line since
y
y
is greater than
−
3
x
2
+
12
-
3
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3 years ago

A 500-gallon tank initially contains 220 gallons of pure distilled water. Brine containing 5 pounds of salt per gallon flows int

Wittaler [7]

Answer: The amount of salt in the tank after 8 minutes is 36.52 pounds.

Step-by-step explanation:

Salt in the tank is modelled by the Principle of Mass Conservation, which states:

(Salt mass rate per unit time to the tank) - (Salt mass per unit time from the tank) = (Salt accumulation rate of the tank)

Flow is measured as the product of salt concentration and flow. A well stirred mixture means that salt concentrations within tank and in the output mass flow are the same. Inflow salt concentration remains constant. Hence:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = \frac{d(V_{tank}(t) \cdot c(t))}{dt}$

By expanding the previous equation:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt} + \frac{dV_{tank}(t)}{dt} \cdot c(t)$

The tank capacity and capacity rate of change given in gallons and gallons per minute are, respectivelly:

$V_{tank} = 220\\\frac{dV_{tank}(t)}{dt} = 0$

Since there is no accumulation within the tank, expression is simplified to this:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt}$

By rearranging the expression, it is noticed the presence of a First-Order Non-Homogeneous Linear Ordinary Differential Equation:

$V_{tank} \cdot \frac{dc(t)}{dt} + f_{out} \cdot c(t) = c_0 \cdot f_{in}$ , where $c(0) = 0 \frac{pounds}{gallon}$ .

$\frac{dc(t)}{dt} + \frac{f_{out}}{V_{tank}} \cdot c(t) = \frac{c_0}{V_{tank}} \cdot f_{in}$

The solution of this equation is:

$c(t) = \frac{c_{0}}{f_{out}} \cdot ({1-e^{-\frac{f_{out}}{V_{tank}}\cdot t }})$

The salt concentration after 8 minutes is:

$c(8) = 0.166 \frac{pounds}{gallon}$

The instantaneous amount of salt in the tank is:

$m_{salt} = (0.166 \frac{pounds}{gallon}) \cdot (220 gallons)\\m_{salt} = 36.52 pounds$

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3 years ago