Alan

The sum of the first eight terms of a particular arithmetic sequence is three times the sum of the first four terms of the sequence. What is the ratio of the first term to the second term? Express your answer as a common fraction.

x = t - 3                    (1)

y = 2*t² - 1               (2)

Rearrange (1):   t = x + 3       (3)

Substitute this into (2):   y = 2*(x+3)² - 1   or   y = 2x² + 12x +17

To find the interval for x, put -2 and 2 into equation (3).

Let v be speed of current.

Let t be time to travel 4 miles downstream.

time = distance/speed

t = 4/(10 + v).    (1)

t + (10/60) = 4/(10 - v).    (2)

You now have two equations in two unknowns.  Solve them to find v.

Note that x = r.cos(theta) and y = r.sin(theta)

Multiply both sides by the denominator of the RHS, and distribute the r.

Can you take it from here?
 

See the answer at https://web2.0calc.com/questions/help-please_53211#r2 

1.  From symmetry, the minimum occurs when OA = OB, where O is the origin.  The tangent line (touching the circle at T, say)  is then at an angle of 45degrees to the horizontal, and 90 degrees to the radial line OT, so forms 45/45/90 right angled triangles with horizontal and vertical axes. This means TA and TB are the same length as OT, which makes AB = 2 

2. Again, symmetry dictates that there is nothing to distinguish a, b, c and d from each other, nor e, f, g and h from each other, so we have :  

       \(a^4=4\\e^4=9\)

   and \((ae)^2+(bf)^2+(cg)^2+(dh)^2=4.a^2.e^2=4.2.3=24\)

The sides of the squares will be 1/4 the perimeters.  The diagonals of the squares will be √2 times the sides.  The diameters of the circles will be the same as the diagonals of the squares.  All that being said, all we need to know is that areas scale as the square of the linear dimensions, so:

Area circle 1/Area circle 2 = (24√2/6√2)² = 16. 

As follows:

I should have written that the horizontal asymptote is y = 4/3 of course, not x!

I get 10 different pairs:

This diagram should help: