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Math help
posted in Off Topic
1
#1
0 Frags +

I can't figure out the equation for the shape, all help is appreciated
http://imgur.com/ASWcEkA

I can't figure out the equation for the shape, all help is appreciated
http://imgur.com/ASWcEkA
2
#2
-12 Frags +

21 xD

21 xD
3
#3
3 Frags +

PQ = ST = x
QR = RS = y

You know that 2x+2y = 30 and the area of the deck is x*x - (x-y)*(x-y), shouldn't be too hard to find the maximum value.

The answer is x = 10, y = 5

PQ = ST = x
QR = RS = y

You know that 2x+2y = 30 and the area of the deck is x*x - (x-y)*(x-y), shouldn't be too hard to find the maximum value.

The answer is x = 10, y = 5
4
#4
5 Frags +
BonkersPQ = ST = x
QR = RS = y

You know that 2x+2y = 30 and the area of the deck is x*x - (x-y)*(x-y), shouldn't be too hard to find the maximum value.

The answer is x = 10, y = 5

You inverted x and y. The sum is right if you give PQ=ST= y and QR = RS = x
Nothing really wrong ah, i just wanted to clarify so that he doesn't confuse

[quote=Bonkers]PQ = ST = x
QR = RS = y

You know that 2x+2y = 30 and the area of the deck is x*x - (x-y)*(x-y), shouldn't be too hard to find the maximum value.

The answer is x = 10, y = 5[/quote]
You inverted x and y. The sum is right if you give PQ=ST= y and QR = RS = x
Nothing really wrong ah, i just wanted to clarify so that he doesn't confuse
5
#5
0 Frags +

Thanks so much everyone

Thanks so much everyone
6
#6
0 Frags +

I am now stuck on two other problems why would they make a 9th grader do this stuff
http://m.imgur.com/s0yfmuh
this is what I have so far for #10
http://m.imgur.com/pFOaWEL

I am now stuck on two other problems why would they make a 9th grader do this stuff
http://m.imgur.com/s0yfmuh
this is what I have so far for #10
http://m.imgur.com/pFOaWEL
7
#7
3 Frags +

Hint for #10:
Since it's an equilateral triangle, all the angles are 60 degrees.

SO you will need to find the area of 1/6 of a circle with radius LC (which you already know how to do), which would be equal to the triangle + 1 shaded region.

Then you can subtract the area of the triangle, and you will be left with the area of 1 shaded region.

Since all 3 shaded regions are identical, multiple the result by 3

you can combine all that into an algebraic eqn and simplify, or just do it step by step (not sure what your teacher wants)

edit: typo

Hint for #10:
Since it's an equilateral triangle, all the angles are 60 degrees.

SO you will need to find the area of 1/6 of a circle with radius LC (which you already know how to do), which would be equal to the triangle + 1 shaded region.

Then you can subtract the area of the triangle, and you will be left with the area of 1 shaded region.

Since all 3 shaded regions are identical, multiple the result by 3

you can combine all that into an algebraic eqn and simplify, or just do it step by step (not sure what your teacher wants)

edit: typo
8
#8
2 Frags +

For the first problem, you can simplify the downward walk to an upward walk that's half as long.
Write fractions with x and y as speed of the slow person and fast person.
You'll end up 735y=630x, which simplifies to the girl being 7/6x as fast.
Now, all you have to do is multiply 1050 (the whole walk since 700 meters+350 meters at the same speed) by 6/7 and you'll find out how far the slower guy made it, assuming up and down are at the same speed. Subtract 700 to be left only with the downward meters, and multiply those by 2, since we earlier disregarded the doubled speed on the way down. Add back the 700. Now you have the total distance the guy walked. Subtract it from 1400 (the distance the girl walked after completing the training).

Edit: If you read it, disregard everything I wrote about chords and shit, I misread the question lol

For the first problem, you can simplify the downward walk to an upward walk that's half as long.
Write fractions with x and y as speed of the slow person and fast person.
You'll end up 735y=630x, which simplifies to the girl being 7/6x as fast.
Now, all you have to do is multiply 1050 (the whole walk since 700 meters+350 meters at the same speed) by 6/7 and you'll find out how far the slower guy made it, assuming up and down are at the same speed. Subtract 700 to be left only with the downward meters, and multiply those by 2, since we earlier disregarded the doubled speed on the way down. Add back the 700. Now you have the total distance the guy walked. Subtract it from 1400 (the distance the girl walked after completing the training).

Edit: If you read it, disregard everything I wrote about chords and shit, I misread the question lol
9
#9
3 Frags +

edit: he removed that part too lol

edit: he removed that part too lol
10
#10
1 Frags +

Ye I typed the stuff about 9) first and somehow ended up thinking you were given the area of the circle and were supposed to calculate the length of that line, and I looked at the imgur and it opened so big I didn't see the text lmao.
(Edit: You wouldn't even need it for that, just a brainfart I guess)
Anyway loot speaks the truth, it's just 1/6th of the circle minus the triangle every time.

Ye I typed the stuff about 9) first and somehow ended up thinking you were given the area of the circle and were supposed to calculate the length of that line, and I looked at the imgur and it opened so big I didn't see the text lmao.
(Edit: You wouldn't even need it for that, just a brainfart I guess)
Anyway loot speaks the truth, it's just 1/6th of the circle minus the triangle every time.
11
#11
4 Frags +

I love this community

I love this community
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