1. How far is the Willis Tower from Navy Pier, given that each unit is approximately 1/8 mile? Round your answer to the nearest tenth of a mile.
The Willis Tower is at the point (-6,-3) and NP is at the point (10,9).
I did this calculation: d^2 = (10 - (-6))^2 + (9 - (-3))^2 = d^2 = 256 + 144 = d = √400 = 20*1/8 = 2.5
Is this correct? also, I know it may be a dumb question, but what would the answer be when I round to the nearest tenth? Would it be 3?
2. Tom is buying a 42in plasma t.v.. The frame around the screen is 1 in wide. Will it fit his entertainment center that has an opening of 36 by 24 inches?
My work:
Okay, so because T.V.s are measured diagonally, we are trying to find the length of a hypotenuse.
24^2 + 36^2 = c^2
576 + 1296 = c^2
c = √1872 = 43.3
Now, because the frame is one inch, wouldn't that ad 1 inch to each corner? Therefore, the length of the hypotenuse in its entirety is 44 inches. Therefore the t,v. would NOT fit in the entertainment center that has an opening of 43.3 inches (diagonally).
Is this correct?
3. I have a box here (it's a rectangle box, not a cube). Its width is 12cm, the length is 9cm and the height is 8cm. I have to find the length of the diagonal line (the hypotenuse of an invisible triangle that goes through it.) Can you help me with this?
Thanks so much!
1 and 2 are correct, as far as I can figure out. You would leave number 1 as 2.5, if not a more complex decimal. As a mathmatician, you will always show the decimals to several points out. Scientists don't like this saying that extra decimals just create confusions. If you have ever heard of the relevant decimals theory, that is the basis. As for 3, try thinking of it as two separate triangles that share a side.
1. To find the distance between the Willis Tower and Navy Pier, we can use the distance formula:
d = √((x2 - x1)^2 + (y2 - y1)^2)
Substituting the given coordinates:
d = √((10 - (-6))^2 + (9 - (-3))^2)
d = √(16^2 + 12^2)
d = √(256 + 144)
d = √400
d = 20 units
Given that each unit is approximately 1/8 mile, the distance is:
20 * (1/8) = 2.5 miles
Rounding to the nearest tenth of a mile, the distance is approximately 2.5 miles.
2. To check if the TV will fit the entertainment center, we can use the Pythagorean theorem:
c^2 = a^2 + b^2
Substituting the given values:
c^2 = 24^2 + 36^2
c^2 = 576 + 1296
c^2 = 1872
c = √1872 ≈ 43.3 inches
However, since the frame is 1 inch wide and it adds to each corner, we need to adjust the width and height of the TV by adding 2 inches to each dimension:
Adjusted width: 36 + 2 = 38 inches
Adjusted height: 24 + 2 = 26 inches
Now, we can re-calculate the diagonal length:
c^2 = 38^2 + 26^2
c^2 = 1444 + 676
c^2 = 2120
c = √2120 ≈ 46 inches
The diagonal length of the TV, including the frame, is approximately 46 inches. Since it is larger than 43.3 inches (the diagonal opening of the entertainment center), the TV will not fit in the entertainment center.
3. To find the length of the diagonal line (hypotenuse) of the box, we can use the Pythagorean theorem:
c^2 = a^2 + b^2
Substituting the given values:
c^2 = 12^2 + 9^2 + 8^2
c^2 = 144 + 81 + 64
c^2 = 289
c = √289 = 17 cm
The length of the diagonal line (hypotenuse) of the box is 17 cm.
1. To find the distance between the Willis Tower and Navy Pier, we can use the distance formula in Cartesian coordinates. The formula is as follows:
d = √[(x2 - x1)^2 + (y2 - y1)^2]
Using the given coordinates:
Willis Tower: (-6, -3)
Navy Pier: (10, 9)
The square of the distance is:
d^2 = (10 - (-6))^2 + (9 - (-3))^2
= (10 + 6)^2 + (9 + 3)^2
= 16^2 + 12^2
= 256 + 144
= 400
Now, taking the square root of 400 gives us:
d = √400 = 20.
Since each unit is approximately 1/8 mile, we need to convert 20 units to miles. 1 mile is equivalent to 8 units.
20 units * (1/8 mile per unit) = 2.5 miles.
So, the distance between the Willis Tower and Navy Pier is approximately 2.5 miles.
2. To determine if the 42-inch plasma TV will fit the entertainment center, we can calculate the diagonal length of the TV, including the frame.
The frame adds 1 inch to each side, so the total dimensions of the TV will be 42 inches + 2 inches (1 inch on each side) = 44 inches.
Using the Pythagorean theorem, we can find the diagonal length:
c^2 = 24^2 + 36^2
= 576 + 1296
= 1872
Taking the square root of 1872 gives us:
c = √1872 ≈ 43.3 inches.
Therefore, the diagonal length of the TV, including the frame, is approximately 43.3 inches.
Since the diagonal length of the TV is greater than the opening of the entertainment center (36 inches), the TV will not fit inside the entertainment center.
3. To find the length of the diagonal line in a rectangular box, we can use the Pythagorean theorem. The Pythagorean theorem states that for a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
In this case, the width is one side (a), the length is the other side (b), and the diagonal line is the hypotenuse (c).
Given that the width is 12 cm, the length is 9 cm, and the height is 8 cm, we can calculate the diagonal length as follows:
c^2 = a^2 + b^2 + h^2
= 12^2 + 9^2 + 8^2
= 144 + 81 + 64
= 289
Taking the square root of 289 gives us:
c = √289 = 17 cm.
Therefore, the length of the diagonal line in the rectangular box is 17 cm.