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PtichkaEL [24]
3 years ago
11

.The spin on a washing machine takes out 35% of the water in the clothes. The clothes in the washing machine contain 3 pints of

water. How much water is left after a spin?
Mathematics
1 answer:
s2008m [1.1K]3 years ago
4 0

Answer: 1.05 Pints

Step-by-step explanation:

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5  and  10

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In the figure below, one side of the right triangle is a diameter of the semicircle.
olga55 [171]

Answer:

Option (A)

Step-by-step explanation:

One side of the given triangle is a diameter of the semicircle given.

Measure of the diameter = 10 units

Total area of the semicircle = \frac{1}{2}\pi (r^{2})

                                              = \frac{1}{2}\pi (5)^2

                                              = 39.27 square units

Area of the right triangle = \frac{1}{2}(\text{Base})(\text{Height})

                                         = \frac{1}{2}(6)(8)

                                         = 24 square units

Area of the shaded region = Area of the semicircle - Area of the right triangle

                                            = 39.27 - 24

                                            = 15.27 square units

                                            ≈ 15 square units

Therefore, option (A) will be the answer.

4 0
3 years ago
Two groups are hiking at about the same speed . Mario’s group hikes for 2.5 hours Seemas group hikes for 4.7 hours and covers 8.
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Answer: 18.8 miles

Step-by-step explanation:

Ww are given the equation 4.7x=2.5x+8.8 to solve the question. This will be:

4.7x = 2.5x + 8.8

Collect like terms.

4.7x - 2.5x = 8.8

2.2x.= 8.8

x = 8.8/2.2

x = 4

To know how fast the kickers are going will be

4.7x = 4.7 × 4 = 18.8 miles

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Towns P,Q,R and S are shown. Q is 35 km due East of P S is 15km due West of P R is 15km due South of P Work out the bearing of R
Anni [7]

Answer:

Part A

The bearing of the point 'R' from 'S' is 225°

Part B

The bearing from R to Q is approximately 293.2°

Step-by-step explanation:

The location of the point 'Q' = 35 km due East of P

The location of the point 'S' = 15 km due West of P

The location of the 'R' = 15 km due south of 'P'

Part A

To work out the distance from 'R' to 'S', we note that the points 'R', 'S', and 'P' form a right triangle, therefore, given that the legs RP and SP are at right angles (point 'S' is due west and point 'R' is due south), we have that the side RS is the hypotenuse side and ∠RPS = 90° and given that \overline{RP} = \overline{SP}, the right triangle ΔRPS is an isosceles right triangle

∴ ∠PRS = ∠PSR = 45°

The bearing of the point 'R' from 'S' measured from the north of 'R' = 180° + 45° = 225°

Part B

∠PRQ = arctan(35/15) ≈ 66.8°

Therefore the bearing  from R to Q = 270 + 90 - 66.8 ≈ 293.2°

6 0
3 years ago
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