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3 years ago

if the focus of an ellipse are (-4,4) and (6,4), then the coordinates of the enter of the ellipsis are

Mathematics

2 answers:

STALIN [3.7K]3 years ago

6 0

I’m sorry but I literally dunno

Ronch [10]3 years ago

3 0

Answer:

The center is (1,4)

Step-by-step explanation:

The coordinates of the center of an ellipse are the coordinates that are in the middle of the two focus.

Then if we have a focus on $(x_1,y_1)$ and $(x_2,y_2)$ , we can say that the coordinates for x and y can be calculated as:

$x=\frac{x_1+x_2}{2}\\ y=\frac{y_1+y_2}{2}$

So, replacing $(x_1,y_1)$ by (-4,4) and $(x_2,y_2)$ by (6,4), we get that the center is:

$x=\frac{-4+6}{2}=1\\ y=\frac{4+4}{2}=4$

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In order for Mark to get to work he needs to drive 10 miles south and 20 miles east. How many miles would he have to go if there

IrinaK [193]

To answer this i need a little more info but it's probably going to be 30 miles.

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2 years ago

a cleaning company charges a flat fee of $8 and $30 per hour to clean a home. write an equation that can represent the situation

Andru [333]

In order to reach the total cost of $128, you need to find the number of hours the company cleaned. The equation to do so would be

30x+ 8 = 128

If you subtract 8 from both sides and then divide by 30 you'll find that x=4.

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3 years ago

Calculate pt3 such that a line from pt1 to pt3 is perpendicular to the line from pt1 to pt2, and the distance between pt1 and pt

Leni [432]

Let the point_1 = p₁ = (1,4)
and      point_2 = p₂ = (-2,1)
and      Point_3 = p₃ = (x,y)

The line from point_1 to point_2 is L₁ and has slope = m₁
The line from point_1 to point_3 is L₂ and has slope = m₂
m₁ = Δy/Δx = (1-4)/(-2-1) = 1
m₂ = Δy/Δx = (y-4)/(x-1)
L₁⊥L₂ ⇒⇒⇒⇒ m₁ * m₂ = -1
∴ (y-4)/(x-1) = -1 ⇒⇒⇒ (y-4)= -(x-1)
(y-4) = (1-x) ⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒ equation (1)

The distance from point_1 to point_2 is d₁
The distance from point_1 to point_3 is d₂
d = $\sqrt{Δx^2+Δy^2}$
d₁ = $\sqrt{(-2-1)^2+(1-4)^2}$
d₂ = $\sqrt{(x-1)^2+(y-4)^2}$
d₁ = d₂
∴ $\sqrt{(-2-1)^2+(1-4)^2} = \sqrt{(x-1)^2+(y-4)^2}$ ⇒⇒ eliminating the root
∴(-2-1)²+(1-4)² = (x-1)²+(y-4)²
(x-1)²+(y-4)² = 18
from equatoin (1) y-4 = 1-x
∴(x-1)²+(1-x)² = 18            ⇒⇒⇒⇒⇒ note: (1-x)² = (x-1)²
2 (x-1)² = 18
(x-1)² = 9
x-1 = $\pm \sqrt{9} = \pm 3$
∴ x = 4 or x = -2
∴ y = 1 or y = 7

Point_3 = (4,1) or (-2,7)

8 0

3 years ago

Find the inverse of ????−7 −2

Masja [62]

Answer:

$A^{-1}=\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}$

Step-by-step explanation:

$A=\begin{bmatrix}-7 & -2\\ 4 & 1\end{bmatrix}$

$det\ A=-7\times 1-(-2)\times 4\\\Rightarrow det\ A=1$

$A^{-1}=\frac{1}{det\ A}\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}\\\Rightarrow A^{-1}=\frac{1}{1}\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}\\\Rightarrow A^{-1}=\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}$

$\therefore A^{-1}=\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}$

$A.A^{-1}=\begin{bmatrix}-7 & -2\\ 4 & 1\end{bmatrix}\times \begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}\\\Rightarrow A.A^{-1}=\begin{pmatrix}\left(-7\right)\cdot \:1+\left(-2\right)\left(-4\right)&\left(-7\right)\cdot \:2+\left(-2\right)\left(-7\right)\\ 4\cdot \:1+1\cdot \left(-4\right)&4\cdot \:2+1\cdot \left(-7\right)\end{pmatrix}\\\Rightarrow A.A^{-1}=\begin{pmatrix}1&0\\ 0&1\end{pmatrix}$