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otez555 [7]
3 years ago
5

My answer was 16, so i thought as a fraction it would 16/100, but i think its 1/6 in the correct way, i dont really know.

Mathematics
1 answer:
larisa [96]3 years ago
4 0

Answer: You're right.

Step-by-step explanation: There are only 6 parts to a die. so the change of landing on 6 would be 1/6, bc there's only 1-6 once.

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755,082 rounded to the nearest 10,000
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760,000 is the answer
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Answer:

B

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8 0
2 years ago
An=3•2^n-1 asap please
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Answer:

This is already in explicit format, there is no more you can do.

Step-by-step explanation:

7 0
2 years ago
Two automobiles left simultaneously from cities A and B heading towards each other and met in 5 hours. The speed of the automobi
quester [9]

Answer:

  450 km

Step-by-step explanation:

<u>Equations</u>

We can define 3 variables: a, b, d. Let "a" and "b" represent the speeds of the cars leaving cities A and B, respectively. Let "d" represent the distance between the two cities. We can write three equations in these three variables:

1. The relation between "a" and "b":

  a = b -10 . . . . . . . the speed of car A is 10 kph less than that of car B

2. The relation between speed and distance when the cars leave at the same time:

  d = (a +b)·5 . . . . . . distance = speed × time

3. Note that the time it takes car B to travel 150 km to the meeting point is (150/b). (time = distance/speed) The total distance covered is ...

  distance covered by car A in 4 1/2 hours + distance covered by both cars (after car B leaves) = total distance

  4.5a + (150/b)(a +b) = d

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<u>Solution</u>

Substituting for d, we have ...

  4.5a + 150/b(a +b) = 5(a +b)

  4.5ab +150a +150b = 5ab +5b^2 . . . . . . multiply by b, eliminate parentheses

  5b^2 +0.5ab -150(a +b) = 0 . . . . . . . . . . subtract the left side

Now, we can substitute for "a" and solve for b.

  5b^2 + 0.5b(b-10) -150(b -10 +b) = 0

  5.5b^2 -5b -300b +1500 = 0 . . . . . . . . eliminate parentheses

  11b^2 -610b +3000 = 0 . . . . . . . . . . . . . multiply by 2

  (11b -60)(b -50) = 0 . . . . . . . . . . . . . . . . factor

The solutions to this equation are ...

  b = 60/11 = 5 5/11 . . . and . . . b = 50

Since b must be greater than 10, the first solution is extraneous, and the values of the variables are ...

  • b = 50
  • a = b-10 = 40
  • d = 5(a+b) = 5(90) = 450

The distance between A and B is 450 km.

_____

<u>Check</u>

<em>When the cars leave at the same time</em>, their speed of closure is the sum of their speeds. They will cover 450 km in ...

  (450 km)/(40 km/h +50 km/h) = 450/90 h = 5 h

__

<em>When car A leaves 4 1/2 hours early</em>, it covers a distance of ...

  (4.5 h)(40 km/h) = 180 km

before car B leaves. The distance remaining to be covered is ...

  450 km - 180 km = 270 km

When car B leaves, the two cars are closing at (40 +50) km/h = 90 km/h, so will cover that 270 km in ...

  (270 km)/(90 km/h) = 3 h

In that time, car B has traveled (3 h)(50 km/h) = 150 km away from city B, as required.

5 0
3 years ago
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Answer:

(7 \times 2 + 8) - ( {x}^{2}  - 2) \\ (14 + 8) - ( {x}^{2}  - 2) \\ 14 + 8 -  {x}^{2}  + 2 \\ 22 +  {x}^{2}  + 2 \\ 24 +  {x}^{2}

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2 years ago
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