A ribbon was cut into 2 pieces whose lengths are in the ratio 3:5. One piece was longer by 4 cm. What are the lengths of the two pieces?
As the ratio is 3:5,Let the lengths be 3x and 5x respectively.Now, by problem-3x+4=5x2x=4x=2Thus, lengths of the pieces are3x= 3*2=6cm and 5x=5*2=10cmAlso, the total length of ribbon is(6+10)cm = 16cm.
What type of rod and reel is best for fishing on the beach?
When it comes to fishing the beach, regardless of my targeted species, I select inexpesnsive reels and moderately priced rods. Sand will eventually get into your gear and all moving parts will be effected. I select gear that I have confidence that I can tear down and thoroughly clean. So mechanically, they are simple devices.My reel & rod choices are gauges for the price/functionality of gear; not pushing any brand or manufacturer.Dependent on the species that I'm targeting, I’ll go with the following:Smaller fish (Perch, Halibut, Striped Bass, Corvina, etc…): Abu-Garcia 6500c/Lamiglas N 386 MC - a lower tier, medium action, two-piece, Stealhead rod. I fish for these species by making a lot of casts with small baits (1/4oz to 3/4oz swimbaits or 1oz to 2oz Carolina rigged grubs/cut bait). I’ll work these baits around troughs, holes, and cuts by constantly casting and retrieving.Bigger fish (Shark, Rays, and large Striped Bass): Penn 500 or 501 Jigmaster/Calstar 270H-8C - a lower tier, 8ft, cork taped (no reel seat), fast action, live-bait rod. When fishing for these species, I'm using a sand spike, chair, and cooler of filled with beer…. I'm casting out cut or whole bait, with an 8oz to 12oz sinker on a sliding rig, backing off the drag, setting the clicker and waiting for things to happen. I rarely fish with this setup. It's mostly the gear I take when a social gathering is more of an emphasis than catching fish.Typically, I'm carrying a couple of spare reels in my tackle/bait bag sealed in zip lock plastic bags. If I accidentally drop my working reel into the sand, I immediately replace it with a spare. Since the reels are somewhat inexpensive, I have spares for spares too.
There is a 40 kg stone that broke into 4 pieces such that you can measure any weight between 1kg to 40 kg using a balance. What are the weights of the broken pieces?
First of all you can measure wt. in the balance in two ways.By putting the weight on one side and the object on the other side.By putting a higher weight or two or more weight altogether on one side and a lower weight or two or more weight s.t. the sum of those is lower than the higher weight along with the object to be measured on the other side. ( I.e. measuring by the difference between the two weights )From my observations I have seen a similarity in these kind of problems.Whatever the total weight. If you have to break it in pieces and then measure from that, in that case you have to find a term X s.t. 3X+1 equals the total weight of the object (in this case 40).So here X is 13. Now write 40 in the form13+(14+13)=40 ~ 13+27=40Now interestingly if you can find a way to measure any weight from 1 to 13 you can measure any weight in the range 40.If you can measure upto 13, you can measure 14 from 27 kg wt (27-13).Then you can measure any weight from 14 to 27 by putting 27 kg in one side and putting the weights in other side from 13 kg to 1 kg.Now even you can measure 27 kg to 40 kg by putting the object on one side and putting the 27 kg combining with the other weights from 1 kg to 13 kg.In similar way you can find how to divide the 13 kg weight in the same pattern.Find x from 3x+1=13 I.e. x=4.Now write 13 in the form4+(5+4)=13 ~ 4+9=13Similarly if you find a way to get the range of 1 to 4 kg you can measure 5 kg (9-4).And then can measure upto 9 kg similarly by substacting and upto 13 kg in same way by addition.To divide 4 kg in same way find x and write 4 like1+(2+1)=4 ~ 1+3=4Now you can see that you can find upto 4 kg by 1 and 3 kg weight.So then we need a 9 kg wt. to measure upto 13 kg and then a 27 kg wt. to measure upto 40 kg.Note: This I found by analysing the problem. There may be a contradiction though. But all I can say that this formulae only works for the wts in the form of 3X+1.
Find the area of a regular nonagon with a perimeter of 144 in.?
Study the links below. The first gives you basics and the second shows you how to determine. With the information you have you'll need to use method four. I am an architect - read graphically oriented right-brain person - so I would recommend drawing these steps out as you follow along, as well as doing this yourself so you understand the process. Nonagon = nine sides/angles, and being regular all sides and angles are equal. Perimeter is the sum of all sides, so each side = 144 in./ 9 sides = 16 inches/side. Next you need to know the angles. Being a regular polygon all sides and angles are equal. Geometry theorem - the sum of the angles = 180 degrees for a triangle, and for each additional side you add another 180 degrees. So start with 180 degrees 9 sides - 3 = 6 x 180 = 1080 + the original 180 = 1,440 degrees / 9 sides = 140 degrees per angle. Next, divide the nonagon into nine pie shaped pieces from the center to each corner, which will give you nine isoceles triangles with the base side at 16 inches and two angles at half the 140 degree angle (because it is a regular polygon) and therefore 70 degrees. Now draw an apothem (a line from the top of your isoceles triangle/center of the nonagon to the midpoint of one side) which will now give you two right triangles as it will be perpendicular to the side. You need to find the length of the apothem with simple trigonometry: Let y = the apothem and x = the known side of the triangle = 8" (half of the 16" that we cut in two with the apothem.) Tangent of 70 degrees = y/x 2.747477419 = y/8 y = 21.97981936 So we have the vertical element of the isoceles triangle, and we can find the area of each, a triangle's area = 1/2 xy = 1/2 (16)(21.97981936 = 175.8385548 sq.in. And we have nine of these: 175.8385548 sq.in. x 9 = 1,582.546994 sq. in. or divide by 144 for square feet = 10.989 or 11 square feet. The second link simplifies this a bit - it gives the formula Area = ap/2 where a = apothem and p = perimeter. I forgot about that one and the above is longhanded/shows the derivation, but I hope will help understand where this comes from.
Homework Question: Joe walked 4 miles north, 9 miles east, 8 miles north, and then 7 miles east. If Joe now decides to walk straight back to where he started, how far must he walk?
The answer provided by User-9597828242429937573 is great, but there are couple of omissions that make its precision unacceptable for practical purposes:1. Our planet is not spherical: its height is less than width due to its rotation. We should take it into account;2. Joe is not a point, but rather a material object. We have to make calculations probably for his geometrical center or, maybe, his center of mass. It should be discussed and decided. If agreed upon geometrical center, 1/2 of Joe's height should be added to the radius;3. Since the journey is long, the result depends on the time he starts and his speed - because of thermal expansion of the Earth's surface heated by the Sun. If Joe is walking while the surface is expanding, his trip back may be longer if the temperature is still high;4. Of course, we should consider the Earth deformation due to the Moon's gravity;5. The answer has to be expressed in closest integer number of Plank lengths. Otherwise, it would be absolutely unrealistic.Other than that, it is pretty accurate and will probably satisfy Joe. :)