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just olya [345]
3 years ago
11

The store manager of Price Chopper wants to put two cereal box displays side by side. Each of the displays will be constructed w

ith forty-eight boxes. The bottom layer of each cereal box display is six cereal boxes long. Each cereal box display cannot be higher than ten boxes. What are two possible sets of dimensions for the two cereal box displays when they are placed side by side? How many cereal boxes can the new display hold?
Mathematics
1 answer:
Volgvan3 years ago
7 0

Answer:

The two possible displays are;

(1) 6 boxes wide by 8 boxes high and

(2) 12 boxes wide and 4 boxes high

The number of cereal box the new display will hold combined is 96 cereal boxes

Step-by-step explanation:

Here we have that the length of the bottom layer = 6 cereal boxes

Maximum height cannot be more than ten boxes high

Number of cereal box per display = 48

Therefore one cereal box display can be 6 boxes wide by 8 boxes high

While if the cereal box dimensions are Length = 2 × Width

Then the second cereal box display can be 12 boxes wide and 4 boxes high

The two possible sets of dimensions are;

(1) 6 boxes wide by 8 boxes high and

(2) 12 boxes wide and 4 boxes high

The first display will hold 48 cereal box displays while the second will hold

12 × 0.5 × 4 × 2 = 48 cereal boxes

The combined number of cereal boxes the new display will hold is 48 + 48 = 96.

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3 0
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while visiting a mountain, John approximated the angle of elevation to the top of a hill to be 25 degrees. Alfter walking 350 fe
strojnjashka [21]
There are several ways of going about this problem, but just know that it all boils down to triangles and the rules/laws of triangles.
So we start by making a large right triangle, because the hill is vertical with the horizontal ground, with 25° as the left angle (where John looks up to top), 90° is the right angle (where ground meets hill base). So we see that the top side of the triangle is the hypotenuse, and equals the line of sight from John to hilltop.
Now we've got additional information that if John walks 350ft towards the hill, his angle of elevation increases by 14. So that = 25+14 = 39. How does that possibly help us?? Well now we can make 2 triangles inside of the one we've already made. So that now we have the triangle base split between the left angle of 25° and where he stopped 350ft to the right of that.
Now the supplement of 39 is 141, and the remaining piece of that too left triangle = 180-25-141 = 14. What does that mean? Well now we have a triangle, where we know all 3 angles and 1 side --> we can find another side by the law of sines:
If a, b and c are the lengths of the legs of a triangle opposite to the angles A, B and C respectively; then the law of sines states:
a÷sinA = b÷sinB = c÷sinC
We really need the hypotenuse to then find our hill height, so we'll make hypotenuse = side b in attached image. That being so, then its opposite angle (B) = 141. And the top right angle (C) = 14 with its opposite side (c) = 350ft.
Now we only need b÷sinB = c÷sinC
So b/sin141 = 350/sin14 --> b = 350sin141/sin14 = 350×.63/
b = 220.3/.24 = 910.47 ft
Now that's our hypotenuse, so using our original large right triangle, we can use right triangular trig. to solve. Let's make the right side, our hill height, equal to x.
Sin ¥ = opp. side / hypotenuse -->
Sin ¥ = x / hypotenuse
Sin25 = x / 910.5
x = 910.5×sin25 = 910.5×.423
x = 384.8 ft

3 0
3 years ago
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