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Dominik [7]
2 years ago
10

The following two sets of parametric functions both represent the same ellipse. Explain the difference between the graphs.

Mathematics
2 answers:
statuscvo [17]2 years ago
7 0
The answer
ellipse main equatin is as follow:

X²/ a²   +  Y²/ b²  =1, where a≠0 and b≠0

for the first equation: <span>x = 3 cos t and y = 8 sin t
</span>we can write <span>x² = 3² cos² t and y² = 8² sin² t
and then  </span>x² /3²= cos² t and y²/8² =  sin² t
therefore,  x² /3²+ y²/8²  =  cos² t + sin² t = 1
equivalent to x² /3²+ y²/8²  = 1

for the second equation, <span>x = 3 cos 4t and y = 8 sin 4t we found
</span>x² /3²+ y²/8²  = cos² 4t + sin² 4t=1

ehidna [41]2 years ago
7 0

Answer with explanation:

The two parametric equation of the same ellipse is

x = 3 cos t and y = 8 sin t

x = 3 cos 4 t and y = 8 sin 4 t

\frac{x}{3}=\cos t\\\\ \frac{x}{3}=\cos 4t\\\\ \frac{y}{8}=\sin t\\\\ \frac{y}{8}=\sin 4t\\\\ (\frac{x}{3})^2+ (\frac{y}{8})^2=\sin^2t+\cos^2t \text{or}=\sin^2 4t+\cos^2 4t=1\\\\\frac{x^2}{9}+\frac{y^2}{64}=1

This is the equation of same ellipse, having different Parametric forms.

→The function involving , 3 cos t and 8 sin t has maximum value, 3 and 8,respectively , and a period of π  , whereas, the function  3 cos 4 t and , 8 sin 4 t , has also same maximum value, 3 and 8,respectively , but period changes , the period after which cycle of trigonometric function  sin 4 t and cos 4 t  repeats is, t=\frac{\pi}{4}.

→x = 3 cos t and y = 8 sin t

\frac{x}{y}=\frac{3}{8 \tan t}\\\\y=\frac{8 \tan t*x}{3}

→x = 3 cos 4 t and y = 8 sin 4 t

\frac{x}{y}=\frac{3}{8 \tan 4 t}\\\\y=\frac{8 \tan 4 t*x}{3}

Also, these are equation of two lines having different slopes both passing through the origin.

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3 0
2 years ago
Choose the mixed number that is equivalent to this decimal 3.242424...
Vinvika [58]

Answer:

3.24 = 3 6/25

Step-by-step explanation:

- Rewrite the decimal number as a fraction with 1 in the denominator.

- Multiply to remove 2 decimal places. Here, you multiply top and bottom by  100

- Find the Greatest Common Factor (GCF) of 324 and 100, if it exists, and reduce the fraction by dividing both numerator and denominator by GCF = 4

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2 years ago
An energy drink container in the shape of a right circular cylinder must have a volume of 19 fluid ounces (1 fluid ounce is appr
omeli [17]

Answer:

the dimension that will minimize cost are radius of 2.79 inch and height of cylinder of 1.40 inch.

Step-by-step explanation:

volume of a circular cylinder = πr²h

volume of 19 fluid ounces (1 fluid ounce is approximately 1.80469 cubic inches) = 19* 1.80469 =34.28911 in³

The cost per square inch of constructing the top and bottom is twice the cost of constructing the lateral side, means

surface area of top + bottom = surface area of side

area of circle(top + bottom) = πr²+πr² = 2πr²

area of the side = 2πrh

the cost of top and bottom = twice the cost of side

2πr² = 2( 2πrh) =  4πrh

divide both side by  2πr

r=  4πrh/ 2πr =2h

r = 2h

for volume = πr²h = 34.28911 in³

and r = 2h

π(2h)²h =34.28911 in³

4πh³ = 34.28911 in³

h³ = 34.28911 /(4*3.142) = 2.7283 in³

h = cube root of 2.7283 in³ = 1.397 inch

r = 2*  1.76 = 2.794 inch

the dimension that will minimize cost are radius of 2.79 inch and height of cylinder of 1.40 inch.

5 0
2 years ago
In the figure shown, FG is tangent to circle D.
romanna [79]

Answer:

Option B

Step-by-step explanation:

From the figure attached,

Circle D is drawn with the radius = DG or DE

A tangent FG has been drawn at a point G on the circle from an external point F.

By theorem,

Radius of a circle is always perpendicular to the tangent, drawn to the circle from an external point.

Therefore, DG ⊥ FG.

Option B will be the correct option.

8 0
2 years ago
Triangle H N K is shown. Angle H N K is 90 degrees. The length of hypotenuse H K is n, the length of H N is 12, and the length o
Mazyrski [523]

Answer:

13

Step-by-step explanation:

From the question, we are given a triangle HNK with an angle of 90°

The length of hypotenuse H K is n,

the length of HN is 12

the length of N K is 6.

From the above values, obtained in the question, we can see that this is a right angled triangle.

We are asked to find the length of the hypotenuse.

We can use Pythagoras Theorem of solve for this.

c² = a² + b²

where c = HK = n

a = NK = 6

b = HN = 12

c² = 6² + 12²

c² = 36 + 144

c² = 180

c = √180

c = 13.416407865

Approximately to the nearest whole number = 13

Therefore the value of HK = n = 13

We can also use Law of Cosines as given in the question to solve for this.

a² = b² + c² - 2ac × Cos A

where c = HK = n

a = NK = 6

b = HN = 12

Hence

c² = a² + b² - 2ab × Cos C

c = √ (a² + b² - 2ab × Cos C)

Where C = 90

c = √ 6² + 12² - 2 × 6 × 12 × Cos 90

c = 13.42

Approximately to the nearest whole number ≈ 13

Therefore the value of HK = n = 13

8 0
2 years ago
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