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AfilCa [17]
3 years ago
5

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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f(a+2)=3a+11/a+2

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2 years ago
F(x)=x^3+4x^2+x-6<br> What are the real zeros
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Happy New Year from MrBillDoesMath!

Answer:

   1, -2, -3

Discussion:

The real roots of the polynomials are factors of the last term (6) divided by the first term (1). The factors of 6/1 or 6 are 1,2,3, and 6. I found by substitution that x= 1 was a solution, divided the polynomial by x-1, came up with a quadratic (x^2 + 5x + 6), and found the remaining roots via the quadratic formula.


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3 years ago
Mr. Lynch bought some oranges and pears.
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Answer:

11: 12

Step-by-step explanation:

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p= 2x -20 -----(1)

Since Mr Lynch did not give any pears away, the number of pears at the end is the initial number of pears.

p= 24

Substitute p= 24 into (1):

2x -20= 24

2x= 24 +20 <em>(</em><em>+</em><em>2</em><em>0</em><em> </em><em>on </em><em>both </em><em>sides</em><em>)</em>

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6 0
3 years ago
Maria correctly concluded that not enough information was provided to construct the requested proof. What extra information woul
Elena-2011 [213]

Answer:

Hello your question is incomplete attached below is the complete question

answer: RS = ST ( option b )

Step-by-step explanation:

First we consider Δ SCA and Δ TCB

∠ SCA  = ∠ TCB  ,  ∠CAS = ∠CBT = 90°

if we assume  RS = ST then ΔSRT will be termed an Isosceles triangle with its legs : SR = ST. also vertex S bisects the base of the triangle.

Δ SBR ≅ ΔSBT  from here we can deduce that ; BR = BT

From the assumptions what is actually missing to contrast the proof is : RS = ST

3 0
3 years ago
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