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White raven [17]
3 years ago
9

"What is the standard form of the ellipse shown?

Mathematics
2 answers:
vampirchik [111]3 years ago
8 0

The answer is:

D) \frac{x^2}{36} + \frac{y^2}{25} =1

lina2011 [118]3 years ago
4 0
The equation of the ellipse ( in the standard form ) :
x² / a² + y²/ b² = 1
a = 6,  b = 5  ( from a graph )
Answer: D ) x² / 36 + y² / 25 = 1
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2. Jessica's grandparents gave her $2000 for college to put in a savings account until she starts
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Answer:

<em>I </em>=2000(0.075)4= $600

Step-by-step explanation:

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2 years ago
Find the area each sector. Do Not round. Part 1. NO LINKS!!<br><br>​
sladkih [1.3K]

Answer:

\textsf{Area of a sector (angle in degrees)}=\dfrac{\theta}{360 \textdegree}\pi r^2

\textsf{Area of a sector (angle in radians)}=\dfrac12r^2\theta

17)  Given:

  • \theta = 240°
  • r = 16 ft

\textsf{Area of a sector}=\dfrac{240}{360}\pi \cdot 16^2=\dfrac{512}{3}\pi \textsf{ ft}^2

19)  Given:

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  • r = 14 cm

\textsf{Area of a sector}=\dfrac12\cdot14^2 \cdot \dfrac{3\pi}{2}=147 \pi \textsf{ cm}^2

21)  Given:

  • \theta=\dfrac{ \pi}{2}
  • r = 10 mi

\textsf{Area of a sector}=\dfrac12\cdot10^2 \cdot \dfrac{\pi}{2}=25 \pi \textsf{ mi}^2

23)  Given:

  • \theta = 60°
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\textsf{Area of a sector}=\dfrac{60}{360}\pi \cdot 7^2=\dfrac{49}{6}\pi \textsf{ km}^2

3 0
2 years ago
Write a real world problem that can be represented by the expression 5x + 10
Yanka [14]
The initial fee for a gym membership is $10 and for each month of the membership it costs $5. Jem has been going to the gym for 8 months, what is the total cost?
7 0
3 years ago
The length of a rectangle is eight more than twice its width. The perimeter is 96 feet. Find the dimensions of the rectangle
Debora [2.8K]

Answer:

The width  of the rectangle is 14'8",  and the length is 33'4"

Step-by-step explanation:

We're given two pieces of information:

The length is eight more than twice the width:

l = 2w + 4

The perimeter is 96 feet:

p = 96

We also need to apply one more piece of information that is not provided here, and that is the relationship between the perimeter of a rectangle, and it's length and width:

p = 2l + 2w

We can solve for w by plugging the other two values into the last:

p = 2w + 2l\\96 = 2w + 2(2w + 4)\\96 = 2w + 4w + 8\\88 = 6w\\3w = 44\\w = \frac{44}{3}\\w = 14\frac{2}{3}

Now we can find the length by plugging w into the first equation:

l = 2w + 4\\l = 2(14\frac{2}{3}) + 4\\l = 29\frac{1}{3} + 4\\l = 33\frac{1}{3}

One third of  a foot is four inches, so the width is 14'8" and the length is 33'4"

To make sure our answer is correct, we should plug those numbers back into the area equation and see if we're right:

p = 2l + 2w\\p = 2(33\frac{1}{3}) + 2(14\frac{2}{3})\\p = 66\frac{2}{3} + 29\frac{1}{3}\\p = 96

So we know our answer's correct

8 0
2 years ago
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