CPhill

Nice job ND !!....I don't see any labels on those cans...must be "generic" cola.....

Thumbs Up + Points......

Assuming that we want the price to be proportional to the volume...

The volume of a cylinder is given by  pi*r^2 * h    where r is the radius and h is the height

So the volume of the smaller can is given by

pi*(3cm)^2 *(12.5cm)= 353.42 cm^3

And the volume of the larger can is given by

pi*(18cm)^2 *(37.5cm) = 38170.35 cm^3

So...using a ratio...

38170.35/353.42 = x/.35      where x is the price of the larger can = $37.80

Wow!!...that's a pretty pricey can of cola....but...it IS a party, after all!!

You're welcome....

That's Standard Southern American English for  'You're welcome"

Well...she spent 30% + 50% = 80% of her money.....so 2 pounds must represent 20% of the money she started with.....and 20% = 1/5 in fractional form....so if 2 pounds is 1/5 of something ......then 5 times that must be the "whole" something. And 2 * 5 = 10 pounds......and that's what she started with.....

Yep....good job......points and thumbs up !!!

This one's easy......

An equation perpendicular to a line with an equation of y = a   is one that has an equation.....x = b

So, if our perpendicular line goes through (6,1)......the equation of this is just x = 6.

Notice, the graph below..... y=4 is a horizontal line where every point has a y value of 4 - no matter what its x value is.  And the line x=6 is a vertical line where every point has an x value of 6 - no matter its y value   !!!    y = 4 is the red line and x = 6 is the blue

All it lacks is a Foreword, a Table of Contents and an Index   ... !!!

Very Good explanation......points and thumbs up.......

Let CG be a tangent to the circle and CE be a secant cutting the circle

angle EDG = (1/2)arc DBE

angle CED = angle FED = (1/2) minor arc FD

angle EDG = angle CED + angle ECD

angle ECD = angle EDG - angle CED

angle ECD = (1/2)arc DBE - (1/2) minor arc FD

angle ECD = (1/2) (arc DBE - minor arc FD)

OK...I can rock and roll with that.......

The arc between the two tangents = 160

So...using your previous problem involving the difference of two arcs.....we have

50 = (1/2)(160 -x)   multiply both sides by 2

100 = 160 - x       then x = 60

Now that's the measure of the arc between the "top" tangent point and the point where the secant first intercepts the circle. So...the remaining arc in the circle =

360 - 160 - 60 = 140

And the measure of "y" = 1/2 the measure of this arc  = 70.

Factoring by "grouping,"  we have

a^2(a + 1) + 1 (a +1) = 0       Taking out the common factor, a+1, we have

(a+1)(a^2+1) = 0          Setting the first factor to 0, we have

a+1 = 0        Subtract 1 from both sides

a = -1            Setting the second factor to 0  produces a non-real solution

So, the only real number solution is a = -1