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Marysya12 [62]
2 years ago
11

A stack of 150 pieces of paper is 2.5 cm thick how many pieces of paper are in a pile 6.5 CM thick​

Mathematics
1 answer:
umka21 [38]2 years ago
5 0

Find the scale factor of the height:

6.5cm / 2.5cm = 2.6

Multiply the known pages by the scale factor:

150 x 2.6 = 390 pieces of paper.

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Answer:

132cm^2

Step-by-step explanation:

first break the shape up into component shapes. the shape is made of two rectangles and two triangles. the area for a rectangle is A=BH where b is base and h is height. the area of a right triangle is BH/2 where b is the base and h is the height. you are given the base and height of the top rectangle 4x7. To find the area of the triangles and the second rectangle you have to do some algebra. the total base length is 19cm and the top rectangle has a length of 7cm so 19-7 leaves you with 12cm to spare and there is two symmetric triangles on each side so 12/2 is 6. now we know all the measurements and can solve.

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An antenna is atop the roof of a 100 foot building, 10 feet from the the edge, as shown in the figure below. From a point 50 fee
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52.) Five times a number is 45
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54).  x/4 = 10    Multiply each side by  4 .
 
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1).  x -10 = 12 .     Add  10  to each side.

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2 years ago
Urgent!! Will mark brainliest!!
horsena [70]

Answer:

1) x is negative and y is positive ⇒ last answer

2) cotФ = -12/35 ⇒ second answer

3) The right identity is cot²Ф - csc²Ф = -1 ⇒ last answer

Step-by-step explanation:

* For any point (x , y) lies on the terminal side of the angle Ф

 in standard position

* x = cosФ and y = sinФ

- If Ф in the first quadrant, then x , y are positive

∴ All trigonometry functions are positive

- If Ф in the second quadrant, then x is negative , y is positive

∴ sinФ only is positive

- If Ф in the third quadrant, then x is negative , y is negative

∴ tanФ only is positive

- If Ф in the fourth quadrant, then x is positive , y is negative

∴ cosФ only is positive

* Lets solve the problems

∵ Ф = 3π/4 ⇒ (135°)

∴ It lies on the second quadrant

∴ x is negative and y is positive

* Lets revise the reciprocal of sinФ, cosФ and tanФ

- cscФ = 1/sinФ

- secФ = 1/cosФ

- cotФ = 1/tanФ

∵ secФ = -37/12

∴ cosФ = -12/37

∵ π/2 < Ф < π

∴ Ф lies on the second quadrant

∴ cotФ is negative values

∵ tan²Ф = sec²Ф - 1

∵ secФ = -37/12

∴ tan²Ф = (-37/12)² - 1 = 1225/144 ⇒ take√ for both sides

∴ tanФ = ± 35/12

∵ cotФ = ± 12/35

∵ cotФ is negative value

∴ cotФ = -12/35

* In the standard position of the angle Ф the terminal

 of it lies on the unit circle O

- By using Pythagorean theorem

∵ x² + y² = 1

∵ x = cosФ and y = sinФ

∴ cos²Ф + sin²Ф = 1 ⇒ (1)

∴ cos²Ф = 1 - sin²Ф

∴ sin²Ф = 1 - cos²Ф

* Divide (1) by cos²Ф

∴ cos²Ф/cos²Ф + sin²Ф/cos²Ф = 1/cos²Ф

* Remember sin²Ф/cos²Ф = tan²Ф and 1/cos²Ф = sec²Ф

∴ 1 + tan²Ф = sec²Ф ⇒ (2) ⇒ subtract 1 from both sides

∴ tan²Ф = sec²Ф - 1 ⇒ subtract sec²Ф from both sides

∴ tan²Ф - sec²Ф = -1

* Divide (1) by sin²Ф

∴ cos²Ф/sin²Ф + sin²Ф/si²Ф = 1/sin²Ф

* Remember cos²Ф/sin²Ф = cot²Ф and 1/sin²Ф = csc²Ф

∴ cot²Ф + 1 = csc²Ф ⇒ (3) ⇒ subtract 1 from both sides

∴ cot²Ф = csc²Ф - 1 ⇒ subtract csc²Ф from both sides

∴ cot²Ф - csc²Ф = -1

* The right identity is cot²Ф - csc²Ф = -1

3 0
3 years ago
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