I have about 5 homework problem that I am stuck on! please help me.
The first of 5 is: A rectangular garden is 20 ft longer than it is wide. Its area is 8000 ft{}^2. What are its dimensions?
Im not even sure where to start with this one.
Next #2:
The area of a rectangle is 16, and its diagonal is \sqrt{68}. Find its dimensions and perimeter.
(x^2)+(y^2)=\sqrt{68}
(x^2)+(y^2)=8.246
xy=16
y=16/x
(x^2)+(16/x)^2=8.246
(x^4)+(96/x)=8.246
Then I got stuck please help me continue.
#3:A box with a square base and no top is to be made from a square piece of carboard by cutting 4 in. squares from each corner and folding up the sides. The box is to hold 784 in{}^3. How big a piece of cardboard is needed?
not sure how to approach this
#4
x^2+9x-1=0
(x-9)x=1
x^2=9x+1
(x-9/2)^2-85/4=0
Solutions x=1/2(-9- sq root 85)
and x=1/2(-9 + sq root 85)
The first portion was marked correctly, however the second solution was marked incorrect. I'm not sure why, please help me figure out why solution #2 is wrong.
Lastly, #5
2x^2+17x+2=0
x^2+17/2 x+ 1=0
x^2+17/2 x= -1
I know to get my final answer I must complete the square but I am not sure how to do so. This is a far as i got before getting stuck.
I've figure out number #1 so I no longer need assistance on that one. But please help me with the following 4 questions.
PLEASE
1. Width = W ft.
Length = (W+20) ft.
A = W*(W+20) = 8,000 Ft^2
W^2 + 20W - 8000 = 0
C=-8,000 = -80*100. Sum = -80+100=20=B.
(w+100)(w-80) = 0
w+100 = 0, W = -100
w-80 = 0, W = 80 Ft = Width.
w+20 = 80+20 = 100 Ft = Length.
2. A = L*W = 16, L = 16/W.
Eq2: L^2 + W^2 = 68
In Eq2, replace L with 16/w:
(16/W)^2 + W^2 = 68
256/W^2 + W^2 = 68
Multiply by W^2:
256 + W^4 = 68W^2
W^4 - 68W^2 + 256 = 0
It was found by trial and error that 2
and -2 satisfied the 4th degree Eq.
W = 2.
L = A/w = 16/2 = 8
P = 2L + 2W = 2*8 + 2*2 = 20
4. x^2 + 9x - 1 = 0
Use Quadratic Formula.
X = 0.10977, and -9.1098
5. 2x^2 + 17x + 2 = 0
Use Quadratic Formula and get:
X = -0.11932, and -8.38068.
1. To solve the first problem, we need to find the dimensions of the rectangular garden. Let's assume the width of the garden is x ft. Since the garden is 20 ft longer than it is wide, the length would be (x + 20) ft.
The area of a rectangle is given by the formula: Area = length × width.
We are given that the area is 8000 ft², so we can set up the equation:
(x + 20) × x = 8000.
Simplify and solve for x:
x² + 20x - 8000 = 0.
Now, you can solve this quadratic equation by factoring (if possible), using the quadratic formula, or by completing the square. Once you find the value of x, substitute it back into the equation to find the length.
2. In the second problem, we are given the area of a rectangle and the diagonal. We need to find the dimensions and perimeter.
Let's assume the length of the rectangle is x and the width is y.
We know that the area is 16, so we have xy = 16.
We are also given the diagonal, which can be determined using the Pythagorean theorem. The diagonal squared is equal to the sum of the squares of the length and width:
x² + y² = (√68)² = 68.
Now you have a system of two equations: xy = 16 and x² + y² = 68. You can solve this system of equations using substitution, elimination, or other methods to find the values of x and y. Once you have the dimensions, you can calculate the perimeter using the formula: Perimeter = 2(x + y).
3. In the third problem, we need to find the size of the cardboard needed to construct a box.
Let's assume the side length of the square base is x inches. If you cut 4 in. squares from each corner, the side of the base will be reduced by 8 in. (2 squares at each end).
So, the base side length will be (x - 8) inches. The height of the box will also be x - 8 inches, as you are folding up the sides.
The volume of a box with a square base is given by the formula: Volume = base area × height.
We are given that the volume is 784 in³, so we can set up the equation:
(x - 8)² × (x - 8) = 784.
Simplify and solve for x:
(x - 8)³ = 784.
Now, solve the equation to find the value of x. Once you have x, you can determine the size of the cardboard needed by adding 8 inches to both dimensions.
4. In problem #4, you have correctly factored the quadratic equation to (x - (9/2))^2 - 85/4 = 0. However, I think you made a mistake while simplifying it further.
(x - (9/2))^2 - 85/4 = 0.
(x^2 - 9x + (81/4)) - 85/4 = 0.
x^2 - 9x + (81/4) - (85/4) = 0.
x^2 - 9x - 4/4 = 0.
x^2 - 9x - 1 = 0.
Now, you can solve this quadratic equation using factoring, the quadratic formula, or completing the square. Once you find the solutions, check your calculations and ensure you've extracted the correct values.
5. For the last problem, you have written x^2 + (17/2)x + 1 = 0. To complete the square, we need to isolate the x terms and move the constant term to the right side of the equation.
x^2 + (17/2)x = -1.
To complete the square, take half of the coefficient of x (which is 17/2), square it, and add it to both sides of the equation.
(x + (17/4))^2 = -1 + (17/4)^2.
Now, simplify the right-hand side and continue solving to find the final answer.