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Write an equation for an ellipse centered at the origin, which has foci at (\pm8,0)(±8,0)left parenthesis, plus minus, 8, comma,

mario62 [17]

Answer:

The equation of the ellipse is $\frac{x^{2}}{(17)^{2}}+\frac{y^{2}}{(15)^{2}}=1$

Step-by-step explanation:

The standard form of the equation of an ellipse with center (0 , 0) is

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ , where

The coordinates of the vertices are (± a , 0)

The coordinates of the foci are (± c , 0) , where c ² = a² - b²

∵ The ellipse is centered at the origin

∴ The center of it is (0 , 0)

∵ It has foci at (± 8 , 0)

- The coordinates of the foci are (± c , 0)

∴ c = ±8

∵ It has Vertices at (± 17 , 0)

- The coordinates of the vertices are (± a , 0)

∴ a = ±17

∵ c² = a² - b²

- Substitute the values of c and a to find b

∴ (8)² = (17)² - b²

∴ 64 = 289 - b²

- Add b² to both sides

∴ b² + 64 = 289

- Subtract 64 from both sides

∴ b² = 225

- Take √ for both sides

∴ b = ± 15

∵ The equation of the ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

- Substitute the values of a and b in it

∴ $\frac{x^{2}}{(17)^{2}}+\frac{y^{2}}{(15)^{2}}=1$ ⇒ $\frac{x^{2}}{289}+\frac{y^{2}}{225}=1$

∴ The equation of the ellipse is $\frac{x^{2}}{(17)^{2}}+\frac{y^{2}}{(15)^{2}}=1$

6 0

3 years ago

What is the Fractional Notation of .0825%?

nirvana33 [79]

Answer:

33/400

Step-by-step explanation:

0.0825=825/10000

=165/2000

=33/400

7 0

3 years ago

I need help with #1 of this problem. It has writings on it because I just looked up the answer because I’m confused but I want t

belka [17]

In the figure below

1) Using the theorem of similar triangles (ΔBXY and ΔBAC),

$\frac{BX}{BA}=\frac{BY}{BC}=\frac{XY}{AC}$

Where

$\begin{gathered} BX=4 \\ BA=5 \\ BY=6 \\ BC\text{ = x} \end{gathered}$

Thus,

$\begin{gathered} \frac{4}{5}=\frac{6}{x} \\ \text{cross}-\text{multiply} \\ 4\times x=6\times5 \\ 4x=30 \\ \text{divide both sides by the coefficient of x, which is 4} \\ \text{thus,} \\ \frac{4x}{4}=\frac{30}{4} \\ x=7.5 \end{gathered}$

thus, BC = 7.5

2) BX = 9, BA = 15, BY = 15, YC = y

In the above diagram,

$\begin{gathered} BC=BY+YC \\ \Rightarrow BC=15\text{ + y} \end{gathered}$

Thus, from the theorem of similar triangles,

$\begin{gathered} \frac{BX}{BA}=\frac{BY}{BC}=\frac{XY}{AC} \\ \frac{9}{15}=\frac{15}{15+y} \end{gathered}$

solving for y, we have

$\begin{gathered} \frac{9}{15}=\frac{15}{15+y} \\ \text{cross}-\text{multiply} \\ 9(15+y)=15(15) \\ \text{open brackets} \\ 135+9y=225 \\ \text{collect like terms} \\ 9y\text{ = 225}-135 \\ 9y=90 \\ \text{divide both sides by the coefficient of y, which is 9} \\ \text{thus,} \\ \frac{9y}{9}=\frac{90}{9} \\ \Rightarrow y=10 \end{gathered}$

thus, YC = 10.

4 0

1 year ago

#1- Jennifer made a box plot to summarize this data. 124, 118, 129, 139, 133, 129, 142, 135, 122, 137. What is the median, the l

svp [43]

Answer:

B or

Lower quartile: 124

Median: 131

Upper quartile: 137

Step-by-step explanation:

To find the median first make the data in order from greatest to least:

118,122,124,129,129,133,135,137,139,142

Then find the middle 2 numbers which are 129 and 133

129+133=262 (divide by 2 to find average) so

median= 262/2=131

To find the Lower quartile find the number in the middle of the first 5 numbers which is 124

To find the Upper quartile find the number in the middle of the second 5 numbers which is 137 so

B or

Lower quartile: 124

Median: 131

Upper quartile: 137

8 0

3 years ago

If a sample mean is 32, which of the following is most likely the range of

Mekhanik [1.2K]

Answer:

Step-by-step explanation:

It is D, I did this quiz once

8 0

3 years ago

Any one for talk by oning video100% safe fna-yzqk-yus. come okaaim arun​​

Any one for talk by oning video100% safe fna-yzqk-yus. come okaaim arun