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lions [1.4K]
3 years ago
8

the base of a lampshade is in the shape of a circle and has diameter of 13 inches. what is the circumference, to the nearest ten

th of an inch of the lampshade? ( use 3.14 for pi)
Mathematics
1 answer:
Semenov [28]3 years ago
5 0
C= 2r x pi c= d x pi 3.14 x 13= 40.82 inches C= 40.8 inches hope this helps
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The solution to 4.2x = 19.32 is x = ___. (3 points)
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Answer:

4.6

Step-by-step explanation:


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Arrange from least to greatest 3,-3, 1, -5, 0, -2, -1, 4, 6.
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-5,-3,-2,-1,0,1,3,4,6

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In a group of EXPLORE students, 38 enjoy video games, 12 enjoy going to the movies and 24 enjoy solving mathematical problems. O
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I really need help with this i beg you !!
sweet-ann [11.9K]

Answer:

The perimeter and area of the square are 56 units and 196 square units, respectively.

Step-by-step explanation:

The inner right triangle represents a 45-45-90 right triangle, which has the feature of a hypotenuse whose length is \sqrt{2} time the length of any of its legs. If the hypotenuse has a measure of 7\sqrt{2}, then the legs of the triangle have a measure of 7.

Now, we are aware that the side length of the square is twice the length of the leg of the right triangle. Then, side length of the square is 14 units long.

Lastly, we know from Geometry that the perimeter and area of the square are represented by the following expressions:

Perimeter

p = 4\cdot l (1)

Area

A = l^{2} (2)

Where l is the side length of the square.

If we know that l = 14, then the perimeter and area of the square are, respectively:

p = 4\cdot (14)

p = 56

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A = 196

The perimeter and area of the square are 56 units and 196 square units, respectively.

7 0
2 years ago
Para embalar una caja se emplea 4,2 m de cinta adhesiva. ?Cuántas cajas se podrán embalar con tres rollos que tienen 3 hm, 7 dam
kompoz [17]

For this case, we perform the conversions:

First roll:

1 \ Hectometer --------> 100 \ meters\\3 \ Hectometer --------> x

x = \frac {3 * 100} {1}\\x = 300 \ meters.

We make a rule of three to determine the number of "c" boxes that can be packed with 300 meters of adhesive tape.

1 -----------> 4.2

c -----------> 300

c = \frac {300 * 1} {4.2}\\c = 71.42857143\\c = 71

You can pack 71 boxes.

Second roll:

1 \ Decametro --------> 10 \ meters\\7 \ Decameter --------> x\\x = \frac {7 * 10} {1}\\x = 70 \ meters.

We make a rule of three to determine the number of "c" boxes that can be packed with 70 meters of adhesive tape.

1 -----------> 4.2

c -----------> 70

c = \frac {70 * 1} {4.2}\\c = 16.66666667\\c = 16

You can pack 16 boxes.

Third roll:

1 -----------> 4.2

c -----------> 50

c = \frac {50 * 1} {4.2}\\c = 11.9047619\\c = 11

You can pack 11 boxes.

Thus, in total you can pack11 + 16 + 71 = 98 \ boxes

Answer:

98 boxes

4 0
3 years ago
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