math help

[netsfan549]

f(x)=2x+8;g(x)= -1x

given f(x) and g(x) find the indicated composition and evaluate

Scott and elizabeth recently built a recangular garden measuring 40 feet by 50 feet. they want to add a paverstone walkway of uniform width around all sides of the garden. How wide can they make the walkway if they purchased enough stones to cover 1000 square feet?

im having trouble with these problems thank you guys

[Patrick Chewing]

You gotta get those x's all one side. That's all I can tell ya. Math is the devil's work.

[DonD13]

do the two problems relate to each other??


for the second one; wouldn't the walkway be 180 feet long? that would give a width of 5,5555

[Theoo]

http://i.imgur.com/RpUJoy1.png

[tpols]

f(x)=2x+8;g(x)= -1x

given f(x) and g(x) find the indicated composition and evaluate

Scott and elizabeth recently built a recangular garden measuring 40 feet by 50 feet. they want to add a paverstone walkway of uniform width around all sides of the garden. How wide can they make the walkway if they purchased enough stones to cover 1000 square feet?

im having trouble with these problems thank you guys
The first one plug g(x) into the x in f(x)

(f o g)(x) = 2(-1x) + 8 or
(g o f)(x) = -1(2x + 8)

Im pretty sure..

Second one if theres 40 + 40 + 50 + 50 length on the outside to be covered, wouldnt you divide that by 1000? So 1000/180 should give you how much the width is

[tpols]

http://i.imgur.com/RpUJoy1.png
:lol

[MavsSuperFan]

f(x)=2x+8;g(x)= -1x

given f(x) and g(x) find the indicated composition and evaluate

Scott and elizabeth recently built a recangular garden measuring 40 feet by 50 feet. they want to add a paverstone walkway of uniform width around all sides of the garden. How wide can they make the walkway if they purchased enough stones to cover 1000 square feet?

im having trouble with these problems thank you guys
Man I used to be so good at math in high school.

I got straight As in AP math and physics with zero effort. But I guess stuff fades if you dont use it long enough :( :( :cry:

So take my help with a grain of salt.
since people have helped you with the first one
For the 2nd question first you calculate the total square footage.
40*50= 2000 square feet.

you can only cover 1000 square feet so that means that you have to figure out a rectangular shape with 1000 square feet with the a 4:5 ratio of width to length that will fit inside the garden and this will be the area not covered by the pavestone. The stuff outside this 1000 square feet will be covered.

so the equation would be (take what I say with a grain of salt)

1000 square feet = X * X(4/5)
1000 =(4(X^2))/5
5000 = 4(X^2)
1250 = X^2
X = 35.35533906 feet

Which is 35 feet 4 inches approximately)

If you put that into the equation you would get

1000 square feet = 35.35533906 * 35.35533906(4/5)
1000 = 35.35533906*28.28427125

So the length of the rectangle inside the original rectangle would be 35.35533906 feet in length and 28.28427125 in width

50feet - 35.35533906 = 14.64466094 feet
40 feet - 28.28427125 = 11.71572875 feet

So that is the perimeters of your walk way i believe.

[MavsSuperFan]

Man I used to be so good at math in high school.

I got straight As in AP math and physics with zero effort. But I guess stuff fades if you dont use it long enough :( :( :cry:

So take my help with a grain of salt.
since people have helped you with the first one
For the 2nd question first you calculate the total square footage.
40*50= 2000 square feet.

you can only cover 1000 square feet so that means that you have to figure out a rectangular shape with 1000 square feet with the a 4:5 ratio of width to length that will fit inside the garden and this will be the area not covered by the pavestone. The stuff outside this 1000 square feet will be covered.

so the equation would be (take what I say with a grain of salt)

1000 square feet = X * X(4/5)
1000 =(4(X^2))/5
5000 = 4(X^2)
1250 = X^2
X = 35.35533906 feet

Which is 35 feet 4 inches approximately)

If you put that into the equation you would get

1000 square feet = 35.35533906 * 35.35533906(4/5)
1000 = 35.35533906*28.28427125

So the length of the rectangle inside the original rectangle would be 35.35533906 feet in length and 28.28427125 in width

50feet - 35.35533906 = 14.64466094 feet
40 feet - 28.28427125 = 11.71572875 feet

So that is the perimeters of your walk way i believe.

actually now that I think about it you have to divide the final figures by 2

14.64/2
and 11.71/2

cause the walkway goes around the edge of the garden.

[TheSilentKiller]

no!

first BTE name that came to my mind!!
I am a math major you scoundrel! :mad:

[netsfan549]

Second problem is mutiple choice

62.5 , 5 , 50, 10

lol

[REACTION]

f(x)=2x+8;g(x)= -1x

given f(x) and g(x) find the indicated composition and evaluate

Depends on what composition they're asking for.

f composed with g = (f ∘ g)(x) = f(g(x)) = -2x + 8
g composed with f = (g ∘ f)(x) = g(f(x)) = -2x - 8


Scott and elizabeth recently built a recangular garden measuring 40 feet by 50 feet. they want to add a paverstone walkway of uniform width around all sides of the garden. How wide can they make the walkway if they purchased enough stones to cover 1000 square feet?

im having trouble with these problems thank you guys

http://i.imgur.com/nu9DHrH.jpg

A(walkway) = A(total) - A(Garden)
= xy - 2000 <= 1000

But since the garden size is fixed, y = x + 10

A(walkway) = x(x + 10) - 2000 = x^2 + 10x - 2000 <= 1000
x <= 50
Max x,y: x = 50, y = 60
A(walkway) = 50*60 - 2000 = 1000 sq ft
Width(walkway) = (50 - 40)/2 = (60 - 50)/2 = 5 ft

[netsfan549]

Depends on what composition they're asking for.

f composed with g = (f ∘ g)(x) = f(g(x)) = -2x + 8
g composed with f = (g ∘ f)(x) = g(f(x)) = -2x - 8



http://i.imgur.com/nu9DHrH.jpg

A(walkway) = A(total) - A(Garden)
= xy - 2000 <= 1000

But since the garden size is fixed, y = x + 10

A(walkway) = x(x + 10) - 2000 = x^2 + 10x - 2000 <= 1000
x <= 50
Max x,y: x = 50, y = 60
A(walkway) = 50*60 - 2000 = 1000 sq ft
Width(walkway) = (50 - 40)/2 = (60 - 50)/2 = 5 ft



Thank you sir

[REACTION]

or actually, I should've just solved directly for the width. It's quicker.

(40 + 2w)(50 + 2w) - 2000 <= 1000
4w^2 + 180w <= 1000
w <= 5

[ace23]

Lol are you in 11th grade?

[Jameerthefear]

the first one is really easy. second one is a bit harder tho

[highwhey]

Visit Khan Academy, they cover every topic from basic arithmetic to algebra, calculus and a bit more advanced algebra.

If you ever plan on studying anything science related in college, your algebra will have to be very solid. Calculus for example, is relatively easy (the calculus concepts), it is the algebra aspect that can prove to be the toughest if you don't have a firm grasp of it.

[REACTION]

Also know your trig.

[no pun intended]

fvck optimization

[DCL]

(40 + x)(50 + x) = 3000
2000 + 90x + x^2 = 3000
x^2 +90x = 1000
x=10

half of x = 5 for each side

[RoseCity07]

The only problem I didn't finish on my calculus finals was the optimization problem. It was the section we covered and I didn't get around to doing the homework. I already dislike word problems enough.

First part: f(-x)=-2x+8

[Bandito]

the first one is really easy. second one is a bit harder tho
you don't know sh1t:lol

[ace23]

you don't know sh1t:lol
Really though, any 8th grader should be able to do the first one.

[Jameerthefear]

you don't know sh1t:lol
yes i do :mad: