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3 years ago

11

How can you compare properties of two functions each represented in a different way?

1 answer:

madam [21]3 years ago

4 0

Step-by-step explanation:

there are three ways to compare functions that is Madhavi Kali numerically and graphically latest compared to function numerical letters considered an example of y equals to 3 X + 5 in this example why is a dependent of ex so in other words if we keep on sending the value of x then why will increase as well

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it is the same distance from second base to first base, and from second base to third base. The angle formed by first base, seco

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A 90 degree angle is formed

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3 years ago

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A trumpet has a length of 140cm. The first valve lowers the root note by two semitones (f1/f0 = 2-2/12) and the second valve low

Alexxandr [17]

Answer:

a. 17.145cm

b. 8.325cm

c. 0.846

d. 2.896

Step-by-step explanation:

It is known that frequency is inversely proportional to length.

Case 1: The player presses only the first valve.

f₁/fo = 2^{-2/12} = lo/l₁

given that lo= 140cm.

l₁ = 2^{2/12}lo = 157.145

change in length:

l₁ - lo = 157.145 - 140 = 17.145cm

Case 2: The player presses only the second valve.

f₂/fo = 2^{-1/12} = lo/l₂

l₂ = 2^{1/12}lo = 148.325cm

l₂ - lo= 148.325-140 = 8.325cm

Case 3: The player presses both the first and second valve.

Change in length is the sum of change in length when they are individually pressed.

l₃ - lo = 17.145+8.325 = 25.47cm

l₃ = 25.47cm+lo = 165.47cm

f₃/fo = 2^{-n/12} = lo/l₃

f₃/fo = 2^{-n/12} = 140/{165.47 = 0.846

2^{-n/12} = 0.846\Rightarrow n/12 = -log(0.846)/log(2) = 0.2413

n = 0.2413*12 = 2.896

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3 years ago

A company is constructing an open-top, square-based, rectangular metal tank that will have a volume of 64 ft3. what dimensions y

Doss [256]

To solve this problem you must apply the proccedure shown below:

1. You must use the formula for calculate the surface area and the volume of the tank:

$1) SA=x^{2} +4xy\\ 2) V=x^{2} y\\ 64=x^{2} y$

2. Solve for $y$ in the second equation and substitute it into the first one.Then, you have:

$y=\frac{64}{x^{2}} \\ SA=x^{2} +4x(\frac{64}{x^{2}})$

3. Now, derivate:

$SA'=2x-256x^{-2} \\ 2x-256x^{-2}=0$

4. When you solve for $x$ , you obtain:

$2x^{3}-256=0\\ x=\sqrt[3]{\frac{256}{2}} \\ x=5.03$

5. Now, you must calculate $y$ :

$y=\frac{64}{(5.03)^{2}} \\ y=2.52$

Therefore, the dimensions are:

The sides of the base of the tank: $5.03 feet$

The height of the tank: $2.52 feet$

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To the nearest tenth of a cubic centimeter, what is the volume of<br> the sphere if r = 17 cm?

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