Help me with this problem! Show the work Please?

Help me with this problem! Show the work Please? Topic: Problem solving problems for adults
July 19, 2019 / By Christa
Question: One evening 478 tickets were sold at the local movie theater. The carges for admission were $ .85 for adults and $ .50 for children. The total receipts for the performance was $ 375.50 . How many adults and how many children attended?
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Best Answers: Help me with this problem! Show the work Please?

Audrey Audrey | 8 days ago
a = adult c = children .85a + .50c = $375.50 This says 'number of adult tickets * price of adult ticket' + 'number of children tickets * price of children ticket' = Total ticket price. a + c = 478 This says 'number of adult tickets' + 'number of children tickets' = 478 tickets Rearrange one equation, then plug it in the other equation. a = 478 - c Plug it in to .85a + .50c = 375.50 .85(478 - c) + .50c = 375.50 And Solve for c. That will give you the number of children tickets. Plug that back into either of the original equations and solve for a, and that will give you the number of adults attended. Good luck! Edit: Also, don't forget to plug in the numbers you get for a and c into the original equation to double check!
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Audrey Originally Answered: Can someone please help me with these last 5 math problems? Please show work for each problem so I understand.?
I noticed that all of you questions involved 'completing the square'. I'm gonna show you how to do two of them, so try the others once again, and ask me if you have any problems :) 1. Lets take: r^2 +4/5r = -2 First, take all the terms to one side. r^2 +4/5r +2 = 0 We need to get an expression in the form: (r+a)^2 +b = 0, where a and b are constants. For that, we can use the identity, (x+y)^2 = x^2 + y^2 +2xy Here, take x = r, and y = a, a constant that we have to find. i.e., (r+a)^2 = r^2 + a^2 +2ar Basically we have to introduce a term that allows us to group 'r^2' and 'r' terms as part of a square, and for that, 2ar = 4/5r .........this is the most important step :) or, a = 2/5 Adding and subtracting a^2 from the original equation, r^2 +4/5r + (2/5)^2 - (2/5)^2 +2 = 0 which can be written as, (r+2/5) - 46/25 = 0 So, the number to be added to both sides is basically (2/5)^2. There's your answer. :) 3. Let the length of the rectangle be L, and the width W. From the information in the question, L + W = 40,or W = 40 - L And, L x W = 336 Substituting for w, we get the equation, L(40 - L) = 336 40L - L^2 = 336 L^2 -40L +336 = 0 Now its just a matter of solving the above quadratic equation to get its roots. Let the roots be L1 and L2. By using the method of 'completing the square', we get the equation, (L - 20)^2 = 64 Solving, we get, L - 20 = 8, or L - 20 = -8 (Since the square root of a positive no. may be positive or negative,i.e., (8)^2 = 64 AND (-8)^2 = 64. However, the second equation can be ignored because it gives you a negative length, which isn't possible) i.e. L = 28 Try the others yourself, that's how you learn :). You should be getting these answers though: 2. 9 4. 1/4 5. Use the equations x - y = 7 and xy = 44, and repeat the steps I followed for question 3. The answer is 11.
Audrey Originally Answered: Can someone please help me with these last 5 math problems? Please show work for each problem so I understand.?
(I'll denote your questions (as ordered) by 1,2,3,4, and 5 as reference.) 1) Assuming your "r" is your "R", to complete the square you must first find two EQUAL numbers that add to give the coefficient of r i.e. 4/5 in this case. This is obviously 2/5. Now multiply these two: 4/15. This is the thing you add onto both sides of the equation (to maintain the equality). Now you just factorise out the left hand side (r^2 + 4/5r + 4/15) which will obviously be a square of a common factor (r+2/5)^2, remembering 2/5 is the number we had previously (this is always the case so don't worry if it's not obvious. Just stick that number in there like that ;) ). So we have (r+2/5)^2= -2+4/15. That's basically it (you can simplify -2+4/15 to -26/15) 2) Similar thing here. The two equal numbers that add to give -6 (don't forget the sign as well) is -3, so you add -3*-3=9 on both sides. Your answer will be (x-3)^2=8 3) This is an algebraic problem involving solving simultaneous equations (if you want to look it up). We denote our unknown wanted value, the length, as x (or whatever your favorite letter is). We also don't know it's width, we don't want it but we do need to involve it so we will call that ermm...y. Okay. So now imagine this rectangle of length x and width y. If you wanted to find the area you would do length x width, so this is x*y which we know is 336. That's our first equation xy=336 Also we know length + width= 40 and so x+y=40. That's our simultaneous equations. Now we want x and so we have to try and get one equation from these two that has only one unknown, x. We can do this by finding y in terms of x from one equation and then putting that into the other: y=40-x from the second equation. So now we can write xy=336 equation only in terms of x by replacing y with 40-x which is what we found to be true from the other equation. So x*(40-x)=336. This gives us a quadratic equation: x^2 - 40x +336 = 0 (by expanding out the brackets and rearranging). I assume you know how to solve these (by the quadratic formulae). There will be two answers: x= 20+1/2*sqrt(1600-1344) =20+8=28; x=20-1/2*sqrt(1600-1344)=20-8=12 So the length is either 28 or 12 and this depends on the width i.e. if the length is 28 then the width is 12 (using y=40-x remember?) and if the length is 12, the width must be 28 (all in meters as we were consistent with the units). Obviously the length was just a label (x) but in reality it is always the longest side and so the length is 28 (the largest value that occured from both equations). See, I even gave you the width on top of that ;) 4) You should know how to do this one now I guess. The two equal numbers that add to give -1 (the coefficient of t) are -0.5 and -0.5*-0.5 is 0.25 which is what we add on both sides. So our answer is (t-0.5)^2=4.25 (see how easy it is?) 5) This is actually like number 3 - simultaneous equations. Call these two integers, x and y (what? I love these letters ;) ). Then x-y=7 and x*y=44 (it doesn't really matter which order as x and y are just labels we used. So whether x-y=7 and y-x=7 is just a matter of what YOU want to call each positive integer. Again writing x=7+y from first equation gives us in the second equation (7+y)*y=44. Go ahead and solve it yourself. You should find again two answers (for each positive integer) but you can merge them into one by forgetting the labels. Don't worry about rating me. I'm just doing my (religious) duty (of Islam) which is to help humanity out at times of need. Peace.
Audrey Originally Answered: Can someone please help me with these last 5 math problems? Please show work for each problem so I understand.?
Q1) If the eqn is: r²+4/5r+2=0,when we complete the sq,we get: (r+2/5)²+46/25=0 I think we need to add 4/25 to both sides as (r+2/5)²= r²+4/5r+4/25 Q2) If the eqn is: x²-6x+1=0,when we complete the sq,we get: (x-3)²-8=0 I think we need to add 9 to both sides as(x-3)²=x²-6x+9 Q3) if length is x and width is y, xy=336 -(1) x+y=40 -(2) from (1),y=336/x -(3) subs.(3) into (2) and we have x=28 or 12 obviously the larger of these 2 values will be the length and the smaller will be the width Q4) If the eqn is: t²-t-4=0,when we complete the sq,we get: (t-1/2)²-17/4=0 I think we need to add 1/4 to both sides as (t-1/2)²=t²-t+1/4 Q5) if one integer is a and the other is b, ab=44 -(1) a-b=7 -(2) from (1), a=44/b -(3) subs.(3) into (2) and we have a=11,b=4 choose either of the two numbers

Abbie Abbie
let x= #/adults and y= #/children x+y=478 .85x+.50y=375.50 multiply 2nd equation by 10, first equation by -5 (or -85 if u want to solve for y) -5x-5y=-2390 8.5x+5y=3750 add: 3.5x=1360 x=1360/3.5= approximately $388.57 (are you sure you entered the problem right??) y+(388.57)=478 y=$89.43
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Abbie Originally Answered: Math problem Please show all work. not a homework problem, but i need step by step instructions?
a palindrome is a number or word that reads the same forwards and backwards the next palindromic reading is 63036 the time lapse is of 1 hour and 40 minutes 63036 - 62926 = 110 miles so its is 110 miles divided by 1 hour 40 min 1 hour 40 min is 1.66667 hours so it is 110 miles divided by 1.66667 110/1.66667 = 66 miles per hour
Abbie Originally Answered: Math problem Please show all work. not a homework problem, but i need step by step instructions?
Rewriting formulae isn't difficult once you know the rules. Step 1: work out what you're trying to solve for. You need to get this term on the left hand side of the equals sign (alone!) - so everything has to be rearranged to do this. Step 2: the key rule is that what you do to one side of the equation, you need to do to the other. Start with addition and subtraction, then do any multiplication and division. Example: x + 4 = 0 subtract 4 from each side and the equation now reads x + 4 -4 = 0 - 4 or x = -4. Remember that the sign on the left of the number is what belongs to it (so in the equation x - 3 = 0 you have to add 3 to each side.) Multiplication and division are opposites, and remember when you do this in an equation you need to do it to *each* term. So 3x = y + 6 can be rewritten as 3 * x = y + 6 and then you need to divide each term by 3 so it becomes x = y/3 + 6/3 or x = y/3 + 2. Another way of writing this is x = (y + 6)/3. Step 3. simplify your answer and make sure the term you are solving for ends up on the left. So to address your specific questions: v=lwh solve for h this can be rewritten as v = lw * h Now divide every term by lw v/lw = lwh/lw which can be simplified to v/lw = h Now switch it around so the h is on the left h = v/lw r=2s-8 solve for s first, add 8 to both sides r + 8 = 2s - 8 + 8 r + 8 = 2s Now you need to divide every term by 2 r/2 + 8/2 = 2s/2 Now simplify r/2 + 4 = s Now move the s to the left hand side s = r/2 + 4 m= a+b/2 solve for a the +b/2 can be subtracted of each side in a lump m - b/2 = a + b/2 - b/2 now simplify m - b/2 = a now switch the equation to write the a on the left a = m - b/2 Remember to always show your working as this often forms part of the marking criteria (even a wrong answer may get some points if there is some correct working shown). I hope this has helped! Good luck.

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