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2 years ago

Will a 11.5-inch x 18-inch rectangular tray fit in the box shown? Explain.

Mathematics

1 answer:

pychu [463]2 years ago

3 0

Answer:

Option D

Step-by-step explanation:

From the picture attached,

In the parallelogram ABCD,

Length of diagonals AC and BD are equal.

By the property,

"If the lengths of the diagonals of a parallelogram are equal, parallelogram will be a rectangle".

Therefore ABCD is a rectangle.

That means all interior angles of the given rectangle will be 90°.

Now we apply Pythagoras theorem in right triangle ΔABC,

AC² = AB² + BC²

(22)² = (12)² + (BC)²

484 - 144 = (BC)²

BC = √340

BC = 18.44 inches

Dimensions of the given box are 12 inches by 18.44 inches.

Dimensions of the rectangular tray is 11.5 inches by 18 inches.

Since, dimensions of the box are larger than the dimensions of the tray, tray can be adjusted in the box.

Therefore, Option D will be the answer.

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3 years ago

A standard piece of paper is 0.05 mm thick. Let's imagine taking a piece of paper and folding the paper in half multiple times.

tatyana61 [14]

Answer:

(a) $g(n)=0.05\cdot 2^n$

(b) $g^{-1}(n)=\log_{2}20g(n)$

Step-by-step explanation:

Part A

The paper's thickness = 0.05mm

When the paper is folded, its width doubles (increases by 100%).

The thickness of the paper grows exponentially and can be modeled by the function:

$g(n)=0.05(1+100\%)^n\\\\g(n)=0.05\cdot 2^n$

Part B

$g(n)=0.05\cdot 2^n\\2^n=\dfrac{g(n)}{0.05}\\ 2^n=20g(n)\\$Changing to logarithm form, we have:\\\log_{2}20g(n)=n\\$Therefore:\\g^{-1}(n)=\log_{2}20g(n)$

Part C

If the thickness of the paper, g(n)=384,472,300,000 mm

Then:

$g^{-1}(n)=\log_{2}20g(n)\\g^{-1}(n)=\log_{2}20\times 384,472,300,000\\=\dfrac{\log 20\times 384,472,300,000}{\log 2} \\g^{-1}(n)=42.8 \approx 43\\n=43$

You must fold the paper 43 times to make the folded paper have a thickness that is the same as the distance from the earth to the moon.

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3 years ago