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Lemur [1.5K]
3 years ago
9

A window is being replaced with tinted glass. The plan below shows the design of the window. Each unit length represents 1 foot.

The glass costs $22 per square foot. How much will it cost to replace the glass? Use 3.14 for π.
g790432


The cost to replace the glass of the window is $
Mathematics
1 answer:
dolphi86 [110]3 years ago
4 0

Answer:

Step-by-step explanation:

Let us solve for the total area. In this problem, we solve area 1 (the rectangle in the middle) and the area 2 ( the two half circle on the sides).

Solve for area 1:

Area rec = L * W = 5units * 4 units = 20 squared units

Solve for area 2:

Area cir = pi *r² = 3.14 * (2 units)² = 12.56 squared units

Total area = 20 + 12.56 = 32.56 foot²

Total cost = 32.56 foot² * ($23/foot²) = $ 748.88

Total cost is $748.88.

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A disadvantage of the contention approach for LANs, such as CSMA/CD, is the capacity wasted due to multiple stations attempting
sammy [17]

Answer:

The overview of the given problem is outlined in the following segment on the explanation.

Step-by-step explanation:

The proportion of slots or positions that have been missed due to numerous concurrent transmission incidents can be estimated as follows:

Checking a probability of transmitting becomes "p".

After considering two or even more attempts, we get

Slot fraction wasted,

= [1-no \ attempt \ probability-first \ attempt \ probability-second \ attempt \ probability+...]

On putting the values, we get

= 1-no \ attempt \ probability-[N\times P\times probability \ of \ attempts]

= 1-(1-P)^{N}-N[P(1-P)^{N}]

So that the above seems to be the right answer.

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PLZ HELP!
elixir [45]

Answer:

if 3 and 5 are the dimensions of the base,

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if 5 and 8 are the base

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if 3 and 8 are the base

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3 years ago
The perimeter of a rectangular plot of land is 280 feet. If one of the sides of the rectangular plot is 50 feet in length, what
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4 0
3 years ago
Read 2 more answers
James cut out four parallelograms, the dimensions of which are shown below. Parallelogram 1 length: 12 in. width: 15 in. diagona
MrRa [10]

Answer: The correct options are;

*The quadrilaterals cannot be placed such that each occupies one quarter of the circle because the vertices of parallelogram 1 do not form right angles

**The quadrilaterals cannot be placed such that each occupies one quarter of the circle because the vertices of parallelogram 4 do not form right angles.

Step-by-step explanation: What James is trying to do is quite simple which is, he wants to place four quadrilaterals inside a circle and he wants the vertices to touch one another at the center of the circle without having to overlap.

This is possible and quite simple, provided all the quadrilaterals have right angles (90 degrees). This is because the center of the circle measures 360 degrees and we can only have four vertices placed there without overlapping only if they all measure 90 degrees each (that is, 90 times 4 equals 360).

We can now show whether or not all four parallelograms have right angles by applying the Pythagoras' theorem to each of them. Note that James has cut the shapes in such a way that the hypotenuse (diagonal) and the other two legs have already been given in the question. As a reminder, the  Pythagoras' theorem is given as,

AC² = AB² + BC² Where AC is the hypotenuse (diagonal) and AB and BC are the other two legs. The experiment would now be as follows;

Quadrilateral 1;

20² = 12² + 15²

400 = 144 + 225

400 ≠ 369

Therefore the vertices of parallelogram 1 do not form a right angle

Quadrilateral 2;

34² = 16² + 30²

1156 = 256 + 900

1156 = 1156

Therefore the vertices of parallelogram 2 forms a right angle

Quadrilateral 3;

29² = 20² + 21²

841 = 400 + 441

841 = 841

Therefore the vertices of parallelogram 3 forms a right angle

Quadrilateral 4;

26² = 18² + 20²

676 = 324 + 400

676 ≠ 724

Therefore the vertices of parallelogram 4 do not form a right angle

The results above shows that only two of the parallelograms cut out have right angles (like a proper square or rectangle for instance), while the other two do not have right angles.

Therefore, the correct option are as follows;

The quadrilaterals cannot be placed such that each occupies one quarter of the circle because the vertices of parallelogram 1 do not form right angles.

The quadrilaterals cannot be placed such that each occupies one quarter of the circle because the vertices of parallelogram 4 do not form right angles.

8 0
3 years ago
Read 2 more answers
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