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Topic: **Problem solving methods of teaching****Question:**
1) x=sin5y and x=(10/pi)y
2) x-5=2y and x-4=(y-1)^2
Compute the area of both. We were taught the following:
1)Find intersection points by setting them equal to each other.
2)Determine which graph lies on top.
3)Set up the integrals and evaluate.
I am having trouble with #1 and #2. Any help is appreciated! Thanks in advance.

May 26, 2019 / By Helene

———————————————— Problem 1 ———————————————— You don't have to set them equal to each other, technically. Any method of solving two equations for an intersection would work. In these cases, though, since they are both solved for the same value (x), that would be the easiest kind of substitution to make to find a solution. x = sin(5y) x = (10/π) y sin(5y) = (10/π) y This can be tricky to solve. The easy method would be to observe: The right side has pi in the denominator, which would be cancelled out by the π that is part of y (which makes sense as the input of sine). Cancel out that and there is 10 left. However, sine will never equal 10. Since the value of y has to go into sine and be the output of sine, it seems unlikely that it will be one of the irrational outputs of sine [such as √(3)/2 ] So, most likely, it is 1, 1/2, or 0 sin(5y) = (10/π) y y = 0 works: sin(5·0) = (10/π) · 0 0 = 0 As does y = π/10: sin(5 · π/10) = (10/π) · π/10 sin(π/2) = 1 1 = 1 y = -π/10 also works: sin(5 · -π/10) = (10/π) · -π/10 sin(-π/2) = -1 -1 = -1 The sine wave has no shifting (no value added inside the function), so it starts like normal sine does. When y=0, it starts at the origin going upwards. The other side is a linear equation with slope 10/π. It goes straight from the origin to the top of the first maximum of the sine wave, which puts it below the curve of the sine wave. If you find this difficult to grasp, graph the first key points of the sine wave. The first would be t=0, and then you would want to find out what would make it equal the next key of the graph of the sine wave (π/2). Then plot the other equation at that point, and since it is a straight line, you can kind of see where it is going to intersect fairly easily. (Just don't forget to graph sine in both directions some.) ———————————————— Problem 2 ———————————————— x - 5 = 2y and x - 4 = ( y - 1 )² Solve for a variable here. Solving for x seems much easier. x = 2y + 5 x = ( y - 1 )² + 4 Graph the line and then graph the parabola. When graphing the parabola, you can take advantage of the fact that the parabola is expressed in vertex form: x = a·( y - h )² + k Where (h, k) is the vertex. So you can graph the point (1,4) as the vertex and then get some points on one side and mirror them to the other side of the vertex. Of course, graphing doesn't work well if they don't intersect at nice exact values. (They probably will here though.) Setting them equal to each other: 2y + 5 = ( y - 1 )² + 4 0 = ( y - 1 )² - 1 - 2y 0 = ( y² - 2y + 1 ) - 1 - 2y 0 = y² - 4y 0 = y(y - 4) 0 = y or 0 = y - 4 4 = y So they intersect at y=0 and y=4. If you graph them, it may help determine which is above the other. Alternatively: x = 2y + 5 is a straight line x = ( y - 1 )² + 4 is an upward opening parabola (given that the square part is positive That means that the line is above the parabola between the intersections.

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We'll just have to assume the simplest thing here, as you do not specify the shape. We will for 2. as well. The simplest thing is the shape is a square. The way to find the area of a square is to multiply the height by the base. So, if you know the area, and one of the height or base, divide the area by the height/base that you know to find the one you don't: 1. Area / height = base = (25 m^2) / (5 m) = 5 m. 2. Area / base = (36 m^2) / (3 m) = 12 m 3. Alright, so... every pair of cars has three meters between it? Side by side, there's three meters in between, back of one to the front of the next, there's three meters in between? OK. Each car, except the final "line" of them each orientation will take up a space 6 m long (3m + 3m) and 5.5 m wide (2.5 m + 3 m). Then there's how you orient them. Fronts facing the 36 m side or facing the 50 m side. We'll find both: a. Fronts facing the 36 m side: The fronts take up 5.5 m (the width and the intervening space). 36 m / 5.5 m = 6 with a remainder of 3 m. 3 m is enough to fit one last car facing that side if no one needs walk over by that side (I'm thinking of inside a building, or outside by a fence - the space outside it would only be 0.5 m, not a satisfactory walking room.) so you can put either 6 or 7 lines of cars facing this way, depending on whether you are allowed to put the last one there with just the 0.5 m of space outside it. Along the 50 m side, you have units of 6 m, so 50 m / 6 m = 8 leaving 2 m extra (5 m after the back end of the last car, not just 3 m). No way to shade an extra one in so this way, facing the 36 m side, you can place 6 * 8 = 48, or 7 * 8 = 56, cars. I feel 48 cars fits the spirit and letter of the description. b. Facing the 50 m side: You need 5.5 m for each except for the last, so you can fit 50 m / 5.5 m = 9 cars facing that side with an extra 0.5 m so the last one has a space of 3.5 m before the wall/fence. How many rows like this? You need 6 m for each row except the last so 36 m / 6 m = 6 rows with the last having its full 3 m of walk-way. So 9 * 6 = 54 cars facing them to the 50 m side. So, going as best as I see with the spirit and letter of the description, I see the maximum being 54 cars with them facing the 50 m side. I won't even go into more complex layouts like one row facing that side, then a row facing the 36 m side, then back to the other way, and so on. 4. I HAVE to assume you mean "tile" not "time" or this doesn't make sense. 2 yd * 4 yd = 8 yd^2. Each square yard is 3 ft * 3ft = 9 ft^2. So, 8 yd^2 * 9 ft^2/yd^2 = 72 square feet. Of tile, not time. 5. Alrighty then. The fence/wall is a total of 600 ft * 8 ft = 4,800 ft^2. If every 12 ft section has a window, there are 600 ft / 12 ft = 50 windows. Each window is 2 ft * 2 ft = 4 ft^2 so 50 of them is 4 ft^2 * 50 = 200 ft^2 that certainly won't be painted. So, a total of 4,800 ft^2 - 200 ft^2 that won't be painted leaves you 4,600 ft^2 to paint. Divide that by 250 ft^2 per gallon of paint: 4,600 ft^2 / 250 ft^2/gal = 18.4 gallons. For one side. Depending on how your teacher likes real world answers, he/she may prefer 19 gallons. But 18.4 gallons is what you actually need.

#1 depends upon the shape of the object..triangle? square..etc. #2 is the same thing, what is the shape? Then use the formula to solve, so for example, if the shape was a triangle, you would plug it into the formula for the area of a triangle: A=1/2bh 36=(1/2)(3)h therefore, h=24m **don't use this answer unless the problem says that the object is a triangle #3 Calculate the total area of the field A=lw A=(50m)(36m)=1800 m^2 then, in order to assure that you leave 3 m in between on all sides of the car, add 3 m to the length and width, and figure out the are of an average car. A=(l+3m)(w+3m) A=(3m+3m)(2.5m+3m)=33m Then, to figure out how many cars fit, use an equation, such as A(total)=A(car)*x with x being the # cars that fit. Then solve for x. 1800=33x x=54.54m So, you would probably round down to 54 cars per field. #4 I'm not really sure what you mean by square feet of time, but the total area of the classroom would be 8 square yards A=lw A=(2yd)(4yd) #5 First, find the area of the wall without the windows A=lw A=(600ft)(8ft)=4800 square feet Then, to figure out how many windows there are, divide the length by 12, which is how often a window appears. So, that would mean there are 50 windows. Now, caculate the area of one window. A=s^2 A=2^2=4 ft Then, subtract the area of the windows, multiplied by the # of windows from the area of the total wall. A(total)-A(window)(#windows) 4800-(4)(50)=4600 square feet Finally, divide the total area by the amount one gallon of paint covers. 4600square feet/250 square feet per gallon=18.4 gallons I would assume that you would round up, because you can't buy .4 of a gallon, so the answer would be 19 gallons of paint. Hope I helped. :]

First off, you gotta make sure to say what shape it is. Merely telling us it's area and height won't help. You could be talking about a trapezoid or something... Secondly this is really easy algebra. For the first two, you know the area is A=(bh)/2 You have enough information to solve for either b or h. The 3rd isn't too hard either. Just set up the equation you have an area of A=36•50 Then divide your A by (3•2.5). 4. I'm sorry you can't have "square feet of time"...that's not a math term or ....ANY term for that matter. With that, I quit. Enjoy!

you are going to decide to resolve for x in each of those equations (so it relatively is going to likely be x = (2/5)y, y = 4, and x = (-a million/2)y + 9/4. The graph of those purposes creates a triangle that shows integrating utilising y = f(x) could require 2 areas. discover the element element of intersection of the x = (2/5)y and x = (-a million/2)y + 9/4 with suggestions from placing them equivalent to eachother. this could yield y = 5/2, this implies that the curves intersect at 5/2, which would be used used as a decrease of your integration. you comprehend the different decrease is y = 4. So your setup would be indispensable from y = 5/2 to y = 4 and your purposes would be setup as ((2/5)y) - ((-a million/2)y + 9/4) dy. you may desire to comprehend something in simple terms combine like words, combine, and plug in limits in accordance to FTC II. desire this facilitates, i'm somewhat rusty on my calc besides the incontrovertible fact that it style of feels good.

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Area is simply the space inside of your shape. Perimeter is simply the length of the sides of your shape, added up. The easiest area to find is of a square, simply square one side and you got its area. To find the perimeter of a square, just take on side, and add it to the the other three sides. For example, you have a square and one of its sides has a length of 4cm. Its area would be 16cm(squared), and the perimeter would be 4+4+4+4 = 16cm. For rectangles, its just Length x Width to find the area. Perimeter is just the same. Adding up all the sides. Sorry, I have no idea what grade level you are in of geometry. Therefore, I'm explaining at a 5th grade level.

Area, the basic area of the shape/figure, often found through differing equation based on the shape Perimeter, the length around the shape/figure Now, its quite hard to be sure what type of probelms you mean when you say "area and perimeter word probelms" as those cover a wide range of different topic/classes in math... However, the most basic concept is to turn all values into variables, this way, we can easily find a relationship between the length, width... etc. There is no way other than that I can explain to you what to do unless you want everything concerning area and perimeter word problems... So I suggest you rephrase your question with an example of a "area and perimeter word problem" so that we may better understand what you are asking

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