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kvv77 [185]
3 years ago
12

This sand timer contains 595 g of sand.

Mathematics
2 answers:
bixtya [17]3 years ago
6 0

Answer:

595g ÷ 30mins = 19.8g per min

19.8g per min ÷ 1.6g each mil

= 12.374mil per min rounded

= 12.4mil per min

snow_lady [41]3 years ago
5 0
The first answer

12.4
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Y=x2<br> x=-3 -2 -1 0 1 2 3<br> y=
brilliants [131]

Answer:

see below

Step-by-step explanation:

Y=x^2

x=-3            -2           -1       0       1       2        3

y=(-3)^2    (-2)^2    (-1)^2    0^2    1^2    2^2   3^2

       9           4           1         0        1         4       9

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3 years ago
If y=20 when x=8, whats the value of x when y=-2
lisabon 2012 [21]

Answer:

x=-4/5

Step-by-step explanation:

8/20=x/-2

simplify 8/20 into 2/5

2/5=x/-2

cross product

5*x=-2*2

5x=-4

x=-4/5

8 0
3 years ago
Extra p!<br> Someone help me, Thanks!!
IRISSAK [1]

Answer:

Step-by-step explanation:

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6 0
3 years ago
Read 2 more answers
Find all polar coordinates of point P where P = ordered pair 3 comma negative pi divided by 3 .
jek_recluse [69]

Answer:

The all polar coordinates of P are:

(3 , -π/3) , (3 , 5π/3) , (-3 , 2π/3) , (-3 , -4π/3)

Step-by-step explanation:

* Lets study the polar coordinates of a point

- In polar coordinates there is an infinite number of coordinates

 for a given point.

- The polar coordinates of a point (x , y) is (r , θ), where

  r = √ ( x2 + y2 )

  θ = tan-1 ( y / x )

# Ex: the following four points are all coordinates for the same point.

* (5 , π/3) = (5 , −5π/3) = (−5 , 4π/3) =(−5 , −2π/3)

- These four points only represent the coordinates of the point without  

  rotating more than once

- So the point (r,θ) can be represented by any of the following

  coordinate pairs  (r , θ + 2π n) and (−r , θ + (2n + 1) π), where n is

  any integer.

* Now lets solve the problem

∵ P = (3 , -π/3)

∵ (r , θ + 2πn)

∴ r = 3 an d Ф = -π/3

- let n = 1

∴ P = (3 , -π/3 + 2π)

∴ P = (3 , 5π/3)

∵ P =(3 , -π/3)

∵ P = (-r , θ + (2n + 1) π)

Let n = 0

∴ P = (-3 , -π/3 + (2×0 + 1) π)

∴ P = (-3 , -π/3 + (0 + 1) π)

∴ P = (-3 , -π/3 + π)

∴ P = (-3 , 2π/3)

∵ P =(3 , -π/3)

∵ P = (-r , θ + (2n + 1) π)

Let n = -1

∴ P = (-3 , -π/3 + (2(-1) + 1) π)

∴ P = (-3 , -π/3 + (-2 + 1) π)

∴ P = (-3 , -π/3 + -π) = (-3 , -4π/3)

∴ P = (-3 , -4π/3)

5 0
3 years ago
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