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Vinvika [58]
3 years ago
10

Skip designs tracks for amusement park rides. For a new design, the track will be elliptical. If the ellipse is placed on a larg

e coordinate grid with its center at (0, 0), the equation (the drawing ^) models the path of the track. The units are given in yards. How long is the major axis of the track? Explain how you found the distance.
Mathematics
1 answer:
SVEN [57.7K]3 years ago
6 0
Given:
center (0,0)
the equation given should have been:
x²/2500 + y²/8100 = 1

We need to identify the larger denominator. If it is under x, the ellipse is horizontal. If it is under y, the ellipse is vertical. In this case, 8100 is larger and it is under y so the ellipse is vertical.

(x-h)²/b² + (y-k)²/a² = 1

h and k are the center values which are both 0.
a = length of the semi-major axis
b = length of the semi-minor axis

x²/2500 + y²/8100 = 1
x²/50² + y²/90² = 1

a = 90. It is the semi-major axis. 

length of the major axis is 180yards.

90 x 2 = 180

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37 apples

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just add 5 to 32

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Simplify the expression. <br> -216) + 3(5 – 4 x 2)
masya89 [10]

Answer:

Step-by-step explanation:

If it's (-216):

(-216) + 3(5 - 4 x 2)

(-216) + 3( -3)

(-216) -9

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-2(16) + 3(5 - 4 x 2)

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4 0
3 years ago
6.
dangina [55]

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3w+10>2w+15

Step-by-step explanation:

If Jerry initially releases 10 videos a week and creates an additional 3 each week, he will create a total of 3w+10 videos, where w is the number of weeks. If Martin initially releases 15 videos a week and creates an additional 2 each week, he will create a total of 2w+10 videos, since w is maintained. Therefore, the inequality \fbox{$3w+10>2w+15$} can be used to find who creates more videos, where w is the number of weeks.

7 0
2 years ago
4.) What is the exact value of sinθ when θ lies in Quadrant II and cosθ=−513
dimaraw [331]

Answer:

Part 4) sin(\theta)=\frac{12}{13}

Part 10) The angle of elevation is 40.36\°

Part 11) The angle of depression is 78.61\°

Part 12) arcsin(0.5)=30\°  or arcsin(0.5)=150\°

Part 13) -45\°  or 225\°

Step-by-step explanation:

Part 4) we have that

cos(\theta)=-\frac{5}{13}

The angle theta lies in Quadrant II

so

The sine of angle theta is positive

Remember that

sin^{2}(\theta)+ cos^{2}(\theta)=1

substitute the given value

sin^{2}(\theta)+(-\frac{5}{13})^{2}=1

sin^{2}(\theta)+(\frac{25}{169})=1

sin^{2}(\theta)=1-(\frac{25}{169})  

sin^{2}(\theta)=(\frac{144}{169})

sin(\theta)=\frac{12}{13}

Part 10)

Let

\theta ----> angle of elevation

we know that

tan(\theta)=\frac{85}{100} ----> opposite side angle theta divided by adjacent side angle theta

\theta=arctan(\frac{85}{100})=40.36\°

Part 11)

Let

\theta ----> angle of depression

we know that

sin(\theta)=\frac{5,389-2,405}{3,044} ----> opposite side angle theta divided by hypotenuse

sin(\theta)=\frac{2,984}{3,044}

\theta=arcsin(\frac{2,984}{3,044})=78.61\°

Part 12) What is the exact value of arcsin(0.5)?

Remember that

sin(30\°)=0.5

therefore

arcsin(0.5) -----> has two solutions

arcsin(0.5)=30\° ----> I Quadrant

or

arcsin(0.5)=180\°-30\°=150\° ----> II Quadrant

Part 13) What is the exact value of arcsin(-\frac{\sqrt{2}}{2})

The sine is negative

so

The angle lies in Quadrant III or Quadrant IV

Remember that

sin(45\°)=\frac{\sqrt{2}}{2}

therefore

arcsin(-\frac{\sqrt{2}}{2}) ----> has two solutions

arcsin(-\frac{\sqrt{2}}{2})=-45\° ----> IV Quadrant

or

arcsin(-\frac{\sqrt{2}}{2})=180\°+45\°=225\° ----> III Quadrant

5 0
3 years ago
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