11 ft X 15 ft = 165 sq ft
165 square feet
The size of a half sheet cak pan is 11x15 inches. 325 degrees for 35-40 minutes.
There are 11 cups of batter in an 11x15 pan for a 2 inch layer. I would guess for an 11x13, maybe 10 cups?
The Kodak model C330 camera takes 4 megapixel pictures. That is large enough to print 11x15 sized pictures. The camera also has an all-glass lens with a 3x optical zoom and 5x advanced digital zoom.
1x15=15 2x15=30 3x15=45 4x15=60 5x15=75 6x15=88 7x15=105 8x15=120 9x15=120 10x15=135 11x15=165 12x15=180 13x15=195 14x15=210 15x15=225
I would recommend trying to find a 15 ft wide carpet that way you will have no seams. To figure out the yards you will need, round the measurements up to 11x15 and multiply those which give you 165. Take 165 divided by 9 = 18.333. So you need 19 yards. If your going to have seams and you have patterns this would not apply. You would have to figure out pattern repeat to figure out just how much you need.
For specific information use theUSPS Link: http://www.usps.com/tools/calculatepostage/welcome.htm?from=global&page=0061calculatepostageIn general, you can mail any envelope up to 11x15 to qualify for first class postage, you then are charged by weight.In general, up to 8 sheets of 8.5x11 paper including envelope will require 2 stamps, 9-15 3 stamps.
That depends on the deceleration applied, in which casetime = 2x45 mph/a = 90/awith a (the deceleration) measured in miles per hour per hour to give the time in hours.If you mean the time that would be taken using the stopping distances in the UK Highway Code, then, assuming a constant deceleration (from the moment the brakes are applied after the thinking distance):The stopping distance at 45 mph is 45 ft (thinking distance) + 101.25 ft (braking distance)[braking distance is speed2 / 20 ft = 452 / 20 ft = 101.25 ft]The thinking distance is travelled at 45 mph, giving:think_time = 45 ft / 45 mph= 45 ft /(45 x 5280 ft / 3600 seconds)= 45 x 3600 / (45 x 5280) seconds= 3600/5280 seconds= 15/22 seconds(This time is constant for all the emergency stopping distances given in the Highway Code.) Two equations of motion can be used to find the stopping time knowing the initial speed and distance (the final speed is zero):final_velocity2 = initial_velocity2 + 2 x acceleration x distancefinal_velocity = 0â†’ acceleration = - initial_velocity2 / (2 x distance)distance = initial_velocity x time + 1/2 x acceleration x time2â†’ distance = initial_velocity x time - (1/4 x initial_velocity2 / distance) x time2â†’ time2 - (4 x distance / initial_velocity) x time + 4 x distance2 / initial_velocity2 = 0â†’ (time - 2 x distance/initial_velocity)2 = 0â†’ breaking_time = 2 x distance / initial_velocity= 2 x (101.25 ft) / (45 x 5280 / 3600 ft per sec)= 202.5 x 3600 / (45 x 5280) seconds= 9x15/2x22â†’ total_stopping_time = 15/22 + 9x15/2x22 seconds= 11x15/2x22 seconds= 15/4 seconds= 33/4 seconds= 3.75 seconds.