A) A rectangle has perimeter 64 cm and area 23 cm^2. Solve the following system of equations to find the rectangle's dimensions.
length=23/width
length + width = 32
B) Solve the system of equations
x^2+y^2=1
xy=0.5
A) A rectangle has perimeter 64 cm and area 23 cm^2. Solve the following system of equations to find the rectangle's dimensions.
length=23/width
length + width = 32
Let
L = length
W = width
L=23/W
L+W=32
23/W+W=32
W2-32W+23=0
Solve for the quadraticequation in W.
Find L using L = 23/W.
You should get 0.7 and 31 approx.
B. Solve
x^2+y^2=1
xy=0.5
If you plot the two equations on sqaure paper, you will see a circle (firt equation) and two disjoint curves with asymptotes toward the 4 extremities of the axes. This will give you an idea where the roots will be, namely where they intersect, and that there should be four of them if they cut the circle, zero if they don't, and two (visually) if they are tangential to the circle.
Substitute y=0.5/x into the equation of the circle (with unit radius).
x2 + (0.5/x)2 = 1
substitute u=x2,
u+0.25/u=1
u2 -u + 0.25 = 0
Solving for u, we get
u=0.5 ± 0 (meaning the intersection is tangential)
=0.5
x=±sqrt(0.5)
y=0.5/x=±sqrt(0.5)
A) Ahh, the case of the elusive rectangle! Let's solve this mathematical mystery, shall we?
We know that the formula for the perimeter of a rectangle is 2(length + width). So, we can write the first equation as 2(length + width) = 64.
Now, remember that the area of a rectangle is found by multiplying the length and the width. In this case, the area is given as 23 cm^2, so we can write the second equation as length * width = 23.
But wait! We also know that the length is equal to 23/width. So, we can substitute this value into the second equation to get (23/width) * width = 23.
Simplifying further, we find that 23 = 23, which is always true! This means that any value of width will satisfy the equation, and therefore, the dimensions of the rectangle are infinitely variable. It's a magical shape that can take on any size!
B) Ah, the marvels of algebraic exploration continue! Let's solve this system of equations and see where it takes us, my friend.
We start with the first equation: x^2 + y^2 = 1. Oh, the intrigue of those squared terms! This equation describes a circle with radius 1 centered at the origin. So, any point (x, y) that lies on this circle will satisfy this equation.
Now, let's move on to the second equation: xy = 0.5. Ah, the mystical crossroads of xy! This equation suggests that the product of x and y is equal to 0.5. How peculiar!
But, you see, my friend, there is no point on the circle x^2 + y^2 = 1 where the product of x and y is 0.5. It's like searching for a unicorn in a forest of giggling fairies - an impossible task!
Therefore, this system of equations has no solutions. It's like trying to find the meaning of life in a bowl of alphabet soup. You may find some letters, but the answer remains elusive.
A) To solve the system of equations, we can substitute the value of length from the first equation into the second equation.
Substituting length = 23/width into length + width = 32, we get:
23/width + width = 32
Multiplying both sides by width, we get:
23 + width^2 = 32width
Rearranging the equation, we get:
width^2 - 32width + 23 = 0
We can solve this quadratic equation using factoring, completing the square, or using the quadratic formula. Let's use factoring:
(width - 1)(width - 23) = 0
Setting each factor equal to zero, we get:
width - 1 = 0 or width - 23 = 0
Solving each equation, we find:
width = 1 or width = 23
Now we can plug these values of width back into the equation length + width = 32 to find the corresponding lengths:
For width = 1, length = 32 - 1 = 31.
For width = 23, length = 32 - 23 = 9.
Therefore, the rectangle's dimensions are 1 cm x 31 cm and 23 cm x 9 cm.
B) To solve the system of equations x^2 + y^2 = 1 and xy = 0.5, we can follow these steps:
From the second equation, we can isolate one variable in terms of the other. Let's solve for x:
xy = 0.5
x = 0.5/y
Now substitute this expression for x in the first equation:
(0.5/y)^2 + y^2 = 1
Simplifying, we get:
0.25/y^2 + y^2 = 1
To solve this equation, we can multiply each term by y^2 to eliminate the denominator:
0.25 + y^4 = y^2
Rearranging the equation, we get a quadratic equation:
y^4 - y^2 + 0.25 = 0
Next, we can substitute a new variable u = y^2 to simplify the equation:
u^2 - u + 0.25 = 0
Now we can solve this quadratic equation:
(u - 0.5)^2 = 0
Taking the square root of both sides, we get:
u - 0.5 = 0
u = 0.5
Substituting back u = y^2, we have:
y^2 = 0.5
Taking the square root of both sides, we get:
y = ±√0.5
Therefore, the solutions for y are y = √0.5 and y = -√0.5.
Now we can substitute these values back into x = 0.5/y to find the corresponding values of x:
For y = √0.5, x = 0.5/√0.5 = √0.5/0.5 = √2/2.
For y = -√0.5, x = 0.5/-√0.5 = -√0.5/0.5 = -√2/2.
So the solutions to the system of equations are (x, y) = (√2/2, √0.5) and (x, y) = (-√2/2, -√0.5).
A) To solve the system of equations, we will use the given information about the rectangle's perimeter and area.
Let's call the length of the rectangle L and the width of the rectangle W.
1) We know that the perimeter of a rectangle is given by the formula: P = 2L + 2W. In this case, the perimeter is given as 64 cm, so we can write the equation as:
2L + 2W = 64
2) We also know that the area of a rectangle is given by the formula: A = L * W. In this case, the area is given as 23 cm^2, so we can write the equation as:
L * W = 23
Now, we have a system of two equations with two variables. We can solve it by substitution or elimination method.
First, let's solve for L in terms of W using the fact that length equals 23 divided by the width:
L = 23/W
Substituting this expression for L into the perimeter equation, we get:
2(23/W) + 2W = 64
Now, we can solve this equation for W.
2(23/W) + 2W = 64
46/W + 2W = 64
Multiply both sides of the equation by W to eliminate the denominator:
46 + 2W^2 = 64W
2W^2 - 64W + 46 = 0
We can use the quadratic formula to solve for W:
W = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 2, b = -64, and c = 46.
Plugging in these values, we get:
W = (-(-64) ± √((-64)^2 - 4(2)(46))) / (2(2))
W = (64 ± √(4096 - 368)) / 4
W = (64 ± √3728) / 4
W = (64 ± 61.08) / 4
This gives us two possible values for W:
W1 = (64 + 61.08) / 4 ≈ 31.77 cm
W2 = (64 - 61.08) / 4 ≈ 0.73 cm
Now, let's substitute the two possible values of W back into the length equation to find the corresponding lengths.
For W = 31.77 cm:
L1 = 23 / 31.77 ≈ 0.72 cm
For W = 0.73 cm:
L2 = 23 / 0.73 ≈ 31.51 cm
Therefore, the two possible dimensions of the rectangle are:
1) Length: 0.72 cm, Width: 31.77 cm
2) Length: 31.51 cm, Width: 0.73 cm
B) To solve the system of equations:
x^2 + y^2 = 1
xy = 0.5
We can solve this system by substituting one equation into the other.
From the second equation, we have:
xy = 0.5
Rearrange this equation to solve for y:
y = 0.5/x
Now substitute this value of y into the first equation:
x^2 + (0.5/x)^2 = 1
To simplify this equation, we need to get rid of the fraction:
x^2 + (0.5^2/x^2) = 1
x^4 + 0.25 = x^2
Rearrange this equation to solve for x^2:
x^4 - x^2 + 0.25 = 0
This equation can be factored as:
(x^2 - 0.5)^2 = 0
Taking the square root of both sides, we get:
x^2 - 0.5 = 0
Solving for x, we have two possible solutions:
x1 = √0.5 ≈ 0.707
x2 = -√0.5 ≈ -0.707
Substituting these values into the second equation, we can find the corresponding y values:
For x = 0.707:
y1 = 0.5 / 0.707 ≈ 0.707
For x = -0.707:
y2 = 0.5 / -0.707 ≈ -0.707
So the solutions to the system of equations are:
1) x = 0.707, y = 0.707
2) x = -0.707, y = -0.707