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Curry
 4 years ago
hard question please help me
ABC is a triangle with sides AB = 6m, BC =8m, and AC = 10m. A line k in the plane of the triangle ABC moves along the segment AC at the rate of 1cm per sec. The line starts at A and ends at C and is always perpendicular to AC.
1. How long does it take the line to reach the point B?
2. How long does it take the line to bisect the area of the triangle?
3. What is the area of the region that the line sweeps in its movement after 6min 45 sec?
4.What is the area of the region that the line sweeps at any give moment from the start of its movement?
Curry
 4 years ago
hard question please help me ABC is a triangle with sides AB = 6m, BC =8m, and AC = 10m. A line k in the plane of the triangle ABC moves along the segment AC at the rate of 1cm per sec. The line starts at A and ends at C and is always perpendicular to AC. 1. How long does it take the line to reach the point B? 2. How long does it take the line to bisect the area of the triangle? 3. What is the area of the region that the line sweeps in its movement after 6min 45 sec? 4.What is the area of the region that the line sweeps at any give moment from the start of its movement?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if you can, draw the figure to go along with it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this IS a tough one...

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2yep it was a precalc problem. i solved 1,3,4 and i need 2 now but im not sure if my answer for 3 and 4 are right but i am sure of number 1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the key is in part 4. if you can figure that out, you should be able to do the rest very easily. give me a moment.

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2i got 2/3cm^2 for every 1 centimeter it moves

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2it is wrong cause there is no way that is possible

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry, had to find some paper

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think you have it right, actually. you have a basic 3/4/5 triangle here. when the line k is moving across, it is creating a smaller 3/4/5 triangle because it is perpendicular to AC.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so the area of the small triangle will be the base 1 cm/s times the height 4/3 cm/s divided by 2; multiply by the amount of time passed.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and to get 2), you figure out how big the total triangle is, and then divide by 2 divide by 1 cm/s, i think.

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2hey here are some flaws with our thinking

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2for number 4 that is true only as long as u reach the altitude after that each one cm is going to create a trapezoid

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's why part 1 is important. it tells you when you need to switch over.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if you're in the US, technically i think it's a trapezium, not a trapezoid; since it won't have parallel sides lol

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2im in the the us we just refer to it as trap in my class

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2since it uses the same formula i believe

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i learned (b1 + b2)*h/2. i am very curious as to what b2 will be. however, there's an easier way, and one that's already in place; split it into 2 triangles. the second triangle will have a new area/s equation.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0have you learned about piecewise functions yet?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it's been a long time since i had precalc

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2number 4 doesnt work for 2/3

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2cause the rea has to be greater than any given side

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is this electronic submission? sounds like what 4 is asking for is an equation.

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2it is asking how much area is covered for every centimeter

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i may be misunderstanding you, but the area swept out changes with every centimeter. this is time dependent, not distance dependent... also remember that k is moving at one CENTImeter per second, and the triangle has sides in meters.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok. didn't mean to be condescending or anything..

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2but if it moves 1 centimeter then thats 1 sec after 1 centimeter the 2 legs are 1cent imeter and 4/3 cm which creates an area of 2/3 but the area has to be greater than the sides

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh i see what you're saying now.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that somehow doesn't seem right, though. if you take 3/4/5 and then divide all sides by two to get 1.5/2/2.5, if you multiply 1.5 by 2 and then divide by 2, you get 1.5; am i thinking about this wrong?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0still, i am doing something wrong.

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2see thats my problem, but is that really possible

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2wait u r doing something wrong

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2wait nvmthats the same thing

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok, here's the problem: when you increase the distance along AC by 1, you're not increasing the area by 2/3cm^2, you're actually increasing the *length of k* by 2/2 cm

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2then the area would be even smaller

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that was a "huh, you're right" not a "huh?"

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2such a carzy problem , my teacher solved it 10 min

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2but refused to chare this with us

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i'm dumb. dumb dumb dumb. i think it's going up by 4/3 not 2/3

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2wow no one can answer this

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2wow no one can answer this

razor99
 4 years ago
Best ResponseYou've already chosen the best response.0so u r really happy about it,

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0OK, I'm commenting so that I have this saved for tomorrow. Curry, post if you don't get help while I'm asleep. I'll bring it to the highermath helpers' attention tomorrow.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0currry do you have answers?

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2im sure this is one of those prob lems that only zarkon or satelite can solve i have the answer for 1 which im sure baout for 3 i got 10.7241

Curry
 4 years ago
Best ResponseYou've already chosen the best response.24 and 2 are insanely unthinkable

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2ig u guys figure out the answer pleazse post

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2ig u guys figure out the answer pleazse post

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2check my answer for number 3

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2so number 2 woulb be 500 secsonds

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2so number 2 woulb be 500 secsonds

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2hey fool if u happen to find all the answers plz post it with the WORK. I got these answers plz do check them 1.6min 2.8.3sec 3.10.7241 as the area 4. ?

Curry
 4 years ago
Best ResponseYou've already chosen the best response.2i will check in the morning for answer. its 1am for me so im going to b ed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0as far as i can tell, 6 minutes is correct.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.08.3 seconds can't be right, though.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah, but that's not question 1 haha. question 1 is just asking how long it takes for k to equal the height. *edited for spelling/grammar*

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0taking the hypotenuse as the base.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328083228792:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok, now let's get back to part 2. first, we have to note what we're looking for; we want the time when the area swept out by k is half of the total area. the total area is 6*8/2 = 24, so we want 24/2 = 12. easy part done. we know just by looking that it's going to be past point B. so we can include that area as already accounted for, and just look for the remainder.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hopefully this isn't too confusing, but we have area 1 as follows: dw:1328085095628:dw and we want to figure out how big area 2 needs to be: dw:1328085234534:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0As you already figured out, Curry, and I obtusely (oh, haha! a math joke...sorry i'll stop now) didn't understand, we have a trapezoid as the second area. Area 1 is, of course, 4.8*3.6/2 = 8.64 m^2 Area 2, therefore, needs to be 12  Area1 = 3.36 m^2 we'll set this equal to the equation for the trapezoid, which is (k+B2)*x/2 = 3.36

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we know that k = 4.8 m, so we need to know what B2 is so that we can solve for x. dw:1328085662851:dw this is a little harder than figuring out the growth of the triangle segment.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0one way to think about it is to take the imaginary little triangle at the top and figure out it's growth instead dw:1328085960597:dw for each increment of 1 cm to the right, the small triangle will grow 3/4 cm in height. since we want B2, we'll take \[k  3/4 \times x\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The area of the trapezoid will therefore be (k + (k(3/4)*x)/2 i don't like their equation builder so i'm going to stop using it until i learn it better.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0set that equal to 3.36 m, and we have a quadratic: .375x^2  4.8 x + 3.36 = 0. solve this for x to get 12.06 m and .743 m we can throw out 12.06 because that would put us beyond the end of the original triangle, but .743 is viable. finally, to get the answer for 2, we add .743 to 3.6, and multiply by 100 to get the number of seconds: 434.3 seconds., or 7 minutes 14ish seconds.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i THINK that's 1 and 2 conclusively solved now. @curry sorry for longwinded answer. 3 and 4 seem poorly worded.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if 3 is asking: "what is the total area swept out by k after 6:45", the answer is Area1 = 8.64, Area2 = 2.084 Area1 + Area2 = 10.7241 meters. so looks like he got that part.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So now, we have to ask again what 4) is really looking for... cuz the way i read it, sounds like it wants an equation. the area swept out per centimeter is constantly changing.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If it wants an equation, it's a bit challenging but doable: By sneaky algebraic methods, the area can be described as: Area = (2/3)*x^2 for x <3.6 Area = 8.64 + 4.8*(x3.6)+.375*(x3.6)^2 for x>3.6 The 3.6 in there should shift the equation correctly so that everything is accounted correctly. I think there's an even sneakier way to combine the two equations into one, but i honestly can't remember/figure out how to do it. it's after 4 am where i live. I encourage you to plug in various values for x between 0 and 10 to see if this works correctly! If it wants a flat number, I'm afraid I don't know how to help you... I hope that the rest of this proves useful, and good luck with your assignment!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for reference: x = 0 should give 0, x = 3.6 should give 8.64, x = 4.343 should give about 12, and x = 10 should give 24.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just to add to the conversation: I found an equation for area thats a bit less complicated x = t/100, where t is in sec dw:1328091690260:dw From the 2 similar triangles we can make the following relationship \[\frac{10x}{y} = \frac{8}{10} \rightarrow y = \frac{5(10x)}{4}\] The 3rd side of the triangle then would be \[\frac{3(10x)}{4}\] Area of the triangle \[\rightarrow \frac{1}{2}(10x)*\frac{3(10x)}{4} = \frac{3}{8}(10x)^{2}\] Now total area of large triangle is 24, so area of part the line has passed is the difference for 0<x<3.6 \[A = \frac{2}{3}x^{2}\] for 3.6 < x <10 \[A =24  \frac{3}{8}(10x)^{2}\]
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