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anonymous
 4 years ago
Mickey sold two types of calendars for her club. She sold a total of 64 calendars for $140.50. One type of calendar sold for $2.50 each, and the other sold for $1.75 each.
1. How many of the $2.50 calendars did she sell?
2. How many of the $1.75 calendars did she sell?
anonymous
 4 years ago
Mickey sold two types of calendars for her club. She sold a total of 64 calendars for $140.50. One type of calendar sold for $2.50 each, and the other sold for $1.75 each. 1. How many of the $2.50 calendars did she sell? 2. How many of the $1.75 calendars did she sell?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let us suppose he is sold x number of one type of calendar @ a price of $2.50 so he sold (64x) number of calenders @ a price of $ 1.75 \[$2.50x + $1.75(64x) = $ 140.50\] now solve for x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So would #1 be 38 & #2 be 26?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i dont know i have not calculated it u cross check your answer by plugging value of x in the equation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok. thats the answers i got. & they worked.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0YAY you did it !!!!!!!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Can you help me with a few more?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0With the wind, a jet takes 3 hours to fly 1890 miles. Against wind, it takes 4 hours for the same trip. Find the winds speed and the planes speed. (Let p=planes speed & w= winds speed.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if speed is p+w the plane travels 1890 miles in 3 hours if speed is pw the plane travels 1890 miles in 4 hours now solve it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok, thank you. I can solve it from there. (: Next one: Flying against the wind, an airplane travels 2880 miles in 4.5 hours. Flying with the wind, the airplane can travel the same distance in 4 hours. 1. Find the speed of the plane in calm air. 2. Find the speed of the wind.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0same problem as before

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I just have problems with setting up the equations.

Hero
 4 years ago
Best ResponseYou've already chosen the best response.0By the way, for the previous problem, the simpler equations are: x = Calendar 1 y = Calendar 2 (calendar amount) x + y = 64 (calendar cost) 2.50x + 1.75y = 140.50

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Can you help me with the last one i posted?

Hero
 4 years ago
Best ResponseYou've already chosen the best response.0It makes no sense to work with an equation if you can't explain what the equation means.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0& half the time you dont make since. Sorry, but with all do respect, i dont like you, one bit.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0& you should consider changing your name from HERO, cause your not! sorry.

Hero
 4 years ago
Best ResponseYou've already chosen the best response.0Wow, really? What did I do to you?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Your all the time butting in when someone else is helping me, and you throw in your 2 cents, and half the time you make NO SINCE!

Hero
 4 years ago
Best ResponseYou've already chosen the best response.0Usually people do that to me. Funny that you're saying it about me. I allowed the other person to explain. Then I "butted in" afterward.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well how about you change your name to MR. BUTTIN.

Hero
 4 years ago
Best ResponseYou've already chosen the best response.0Perhaps a visual diagram would help, but understand that both problems are related to each other only because they are systems of equations. Other than that, they are two completely different problems.

Hero
 4 years ago
Best ResponseYou've already chosen the best response.0As far as my explanation, I can take another stab at it to make it super clear. If it isn't clear after that, then I won't bother you anymore.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Look im sorry, i really aint got time to bulll crap around, i got finals Monday for 2 different maths, and im freakin out!

Hero
 4 years ago
Best ResponseYou've already chosen the best response.0It is easier to work with variables then entire expressions, so we usually define x = number of 1st calendar type y = number of 2nd calendar type The problem states that there are a total of 64 calendars, so: Calendar 1 + Calendar 2 = 64   x + y = 64 The problem states that 1 calendar costs 2.50 each. The other costs 1.75 each: Calendar 1 costs 2.50 each Calendar 2 costs 1.75 each The total amount of calendars sold costs 140.50. In other words, when all 64 calendars together are sold the total cost will be 140.50: 2.50(Calendar 1) + 1.75(Calendar 2) = 140.50   2.50 x + 1.75 y = 140.50 So now, all we do is write both equations together as a system of equations: x + y = 64 2.50x + 1.75y = 140.50 Try to keep in mind what x and y means as you observe these equations.

Hero
 4 years ago
Best ResponseYou've already chosen the best response.0Please try to control your frustration. It is counterproductive to studying.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@pamelaa23 hero is right you should try to understand the thing.........but that was hillarious Mr. ButtIn

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Lol, i know. But, i thought after that problem i was caught up on work and everything. & then a friends text me and reminded me about our other work. * turns out its due tomorrow, 75 questions of Trig/Algebra & a bunch more work.
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