Amanda said to Dad, no matter what happens, I know you'll always be with me

Amanda said to Dad, no matter what happens, I know you'll always be with me

Amanda told his father that no matter what happened, he knew his father would always be with him

"The father said to his son," no matter what happens, I'll be with you! "

The father said to his son, no matter what happens, he will always be with him!

If a vertex and two focal points of an ellipse form an equilateral triangle, the eccentricity is?
The answer is 1 / 2

Because one vertex of the ellipse and two focal points form an equilateral triangle
So the root 3C = B
c^2+b^2=a^2
3c^2=b^2
So 4C ^ 2 = a ^ 2
e=c/a=1/2

Skillfully calculate 1.2 + 1.3 + 1.4 +9.7+9.8
1.2+1.3+1.4…… +How to calculate 9.7 + 9.8?

1.2+1.3+1.4…… +9.7+9.8
=（1.2+9.8）+（1.3+9.7）+.+5.5
=11+11+.+5.5
=5.5×87
=478.5;
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It is known that the two of the equations 3x square + 6x-1 = 0 are X1 and X2, and 1 / 1 of X1 + 1 / 2 is equal to 0

The square of 3x + 6x-1 = 0, Weida theorem: X1 + x2 = - B / a = - 2, x1x2 = C / a = - 1 / 31 / X1 + 1 / x2 = (x1 + x2) / x1x2 = - 2 / (- 1 / 3) = the square of 63x + 6x-1 = 0

The focus is on the x-axis, the length of the major axis is twice that of the minor axis, and the standard equation of the ellipse passing through the point (2, - 6) is

x^2/a^2+y^2/b^2=1
a=2b
x^2+4y^2=4b^2
Over (2, - 6) 4 + 4 * 36 = 4B ^ 2
b^2=37
a^2=4b^2=148
x^2/148+y^2/37=1

4.5.9.14.23

an=a[(1+√5)/2]^(n-1)+b[(1-√5)/2]^(n-1)
This is a variant of the Fibonacci sequence
By:
a1=a+b=4
a2=a((1+√5)/2)+b((1-√5)/2)=5
The solution is as follows
a=2+3/√5
b=2-3/√5

How to solve the problem that 2x minus one equals four minus zero 5x

Hello!
2X-1=4-0.5X
2.5x=5
x=2
If it helps you, please accept it

One and three fourths is a fraction

Seven quarters

Let A1 A2 A3 be linearly independent. If β can be expressed linearly by A1 A2 A3, the expression is unique

Let β = p1a1 + p2a2 + p3a3 = q1a1 + q2a2 + q3a3, then (p1-q1) a1 + (p2-q2) A2 + (p3-q3) A3 = 0. From the linear independence of A1, A2, A3, p1-q1 = p2-q2 = p3-q3 = 0