I need help with these problems can someone show me how to solve them step by step.If anyone has time i appreciate it.
Problem #1
Geometry. The length of a rectangle is 1 cm longer than its width. If the diagonal of the rectangle is 4 cm, what are the dimensions (the length and the width) of the rectangle?
Problem#2
Science and medicine. The equation
h=-16t^2+112t gives the height of an arrow, shot upward from the ground with an initial velocity of 112 ft/s, where t is the time after the arrow leaves the ground. Find the time it takes for the arrow to
reach a height of 180 ft.
first:
l-1=w
diagonal^2=length^2 + width^2
16=l^2 + (l+1)^2
solve for l
On the second, put 180 onto the equation , getting
16t^2 +112t - 180=0
or use the quadratic equation
I would divide by four to simplify it.
Sure! I'll explain step by step how to solve each problem.
Problem #1:
To find the dimensions of the rectangle, we'll start by setting up some equations based on the given information.
Let's say the width of the rectangle is "w" cm. Since the length is 1 cm longer than the width, the length can be represented as "w + 1" cm.
We are also told that the diagonal of the rectangle is 4 cm. Using the Pythagorean theorem, we know that the square of the diagonal is equal to the sum of the squares of the length and width.
So, the equation becomes:
(diagonal)^2 = (length)^2 + (width)^2
Substituting the known values:
4^2 = (w + 1)^2 + w^2
Now, let's solve for w, the width of the rectangle.
16 = w^2 + 2w + 1 + w^2
16 = 2w^2 + 2w + 1
Rearranging the equation:
2w^2 + 2w - 15 = 0
Now, we have a quadratic equation in terms of w. We can solve this using factoring or the quadratic formula.
If you want to use the quadratic formula, we'll first divide the equation by 2 to simplify it:
w^2 + w - 15/2 = 0
Now, applying the quadratic formula:
w = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = 1, and c = -15/2. Substitute these values into the formula to find the possible solutions for w. Then, you can find the corresponding length by adding 1 to each value of w.
Problem #2:
We are given an equation that describes the height of an arrow as a function of time after it leaves the ground. The equation is:
h = -16t^2 + 112t
We need to find the time it takes for the arrow to reach a height of 180 ft. To do this, we can set up the equation and solve for t.
Substituting the given height into the equation, we have:
180 = -16t^2 + 112t
Next, we can rearrange the equation to solve for t. Let's move everything to one side:
-16t^2 + 112t - 180 = 0
In this case, we have a quadratic equation in terms of t. Since it doesn't factor easily, using the quadratic formula might be the easiest approach.
Divide the equation by -4 to simplify it:
4t^2 - 28t + 45 = 0
Now, we can apply the quadratic formula to find the possible solutions for t.
Once you have the solutions, you can compare them and select the appropriate value for the time it takes for the arrow to reach a height of 180 ft.
Remember, the quadratic formula is:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 4, b = -28, and c = 45. Substitute these values into the formula to find the possible solutions for t.