Can someone set this up in equations so i can solve them i would greatly appreciate it. THanks.

Problem #3
The base of a ladder is 14 feet away from the wall. The top of the ladder is 17 feet from the floor. Find the length of the ladder to the nearest thousandth.

Problem #4
A rectangular garden is to be surrounded by a walkway of constant width. The garden's dimensions are 30 ft by 40 ft. The total area, garden plus walkway, is to be 1800 ft2. What must be the width of the walkway to the nearest thousandth?

112 because im smart im 10 and im in the 11th grade

oh sorry the answer is 2,000

the next one is 4,000

-x+y=6

I do not want the answers just the equations.

3) Think about what shape that would make. If you're not sure, draw it out. The formula? a^2 + b^2 = .... ?

4) Again, this may help to draw out a picture of it. It would look like a rectangle inside of a rectangle. Now that you see that much, what's the area of the garden? We know it's 30ft x 40ft - which is 1200 ft^2

The area of the whole place is 1800 ft^2

So subtract 1800 - 1200 and we see the area of the walkway is 600 ft^2

Try to figure it out from there and see what you come up with.

Matt

EQUATION:

(30+2x)(40+2x) = 1800 sq ft.
4(15+x)(20+x) = 1800
x^2+35x+300=450
x^2+35x-150=0
Use quadratic formula to get:
x=[-35+sqrt(35^2-4*-150)]/2
x=[-35+-sqrt1825]/2
x=[-35+-42.72]/2
x=7.72/2
x=3.86 ft

Problem 4 solution:

EQUATION:
(30+2x)(40+2x) = 1800 sq ft.
4(15+x)(20+x) = 1800
x^2+35x+300=450
x^2+35x-150=0
Use quadratic formula to get:
x=[-35+sqrt(35^2-4*-150)]/2
x=[-35+-sqrt1825]/2
x=[-35+-42.72]/2
x=7.72/2
x=3.86 ft

Sure! I can help you set up the equations for both problems.

Problem #3:
Let's assume that the length of the ladder is denoted by 'x'. We know that the base of the ladder is 14 feet away from the wall and the top of the ladder is 17 feet from the floor. This forms a right triangle where the ladder is the hypotenuse.

Using the Pythagorean theorem, we have:

x^2 = 14^2 + 17^2

Simplifying this equation gives:

x^2 = 196 + 289
x^2 = 485

To solve for 'x', we take the square root of both sides of the equation:

x = sqrt(485)

So the length of the ladder is approximately equal to the square root of 485.

Problem #4:
Let's assume the width of the walkway is denoted by 'w'. The dimensions of the garden are given as 30 ft by 40 ft. To find the total area, we need to add the area of the garden to the area of the walkway, which is given as 1800 ft^2.

The area of the garden is given by the length multiplied by the width:

Area of garden = 30 * 40 = 1200 ft^2

The area of the walkway can be calculated by subtracting the area of the garden from the total area:

Area of walkway = Total area - Area of garden = 1800 - 1200 = 600 ft^2

To find the width of the walkway, we divide the area of the walkway by the length of the garden plus twice the width of the walkway:

w = Area of walkway / (30 + 2w) + (40 + 2w)

Substituting the value of the area of the walkway, we get:

w = 600 / (30 + 2w) + (40 + 2w)

You can simplify this equation and solve for 'w' to find the width of the walkway.

I hope this helps! Let me know if you have any further questions.