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scoray [572]
3 years ago
13

Smh, what is this. If you answer this, please add a prove it statement. Thank you.​

Mathematics
1 answer:
OlgaM077 [116]3 years ago
6 0
28 slices
10 1/2 divided by 3/8 is 28
28 times 3/8 equals 10 1/2
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1. Kyle and John are twins who decided to plan an imaginary trip that started with cliff jumping. There were two different spots
Vikki [24]

Answer:

a) Δx = -35ft

b) Δx = -25ft

c) Kyle traveled 10ft more than John

Step-by-step explanation:

We define a coordinate reference system() in which y = 0 corresponds to the water surface.

a) The initial position of Kyle in our coordinate reference system is:

   x_{o} = 20 ft

   and his final position is:

   x_{f} = -15 ft

   Therefore, he traveled

   Δx = x_{f} - x_{o} = -15 - 20 = -35 ft

b) The initial position of John in our coordinate reference system is:

   x_{o} = 15 ft

   and his final position is:

   x_{f} = -10 ft

   Therefore, he traveled

   Δx = x_{f} - x_{o} = -10 - 15 = -25 ft

c) Then, Kyle traveled 10ft more than John

 

6 0
3 years ago
(x^3)^9 = <br> A: x^12 <br> B: x^27 <br> C: x^3 <br> D: x^6
Grace [21]
(x^3)^9=x^{27}
3 0
3 years ago
X+12=31 solve for x i’m in a breakout room i have limited time to finish this
Nitella [24]

Answer: x=19

Step-by-step explanation:

x=31-12

x=19

4 0
2 years ago
Click on the area of the squarWhich equation would be used to represent the word problem below?
gogolik [260]

Answer:5.50x + 9 = 2.50x

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Two dice are thrown. Let E be the event that the sum of the dice is
bearhunter [10]

Answer:

  • P(E) = 1/2
  • P(F) = 11/32
  • P(G) = 1/6
  • P(EF) = 5/52
  • P(FG) = 1/32
  • P(EG) = 1/6

Step-by-step explanation:

For the sum to be even, both dice can be odd, or both even. The probability of a dice being odd is 1/2 and the same is for it to be even. Since the result of the dices are independent, we have that

P(E) = (1/2)² + (1/2)² = 1/2

Out of the 36 possible outcomes for the dice (assuming that you can distinguish between first and second dice), there are 11 cases in which one dice is a 6 (if you fix 1 dice as 6, there are 6 possibilities for the other, but you are counting double 6 twice, so you substract one and you get 6+6-1 = 11). Since all configurations for the dices have equal probability, we get that

P(F) = 11/32

The probability for the second dice to be equal to the first one is 1/6 (it has to match the same number the first dice got). Hence

P(G) = 1/6

for EF, you need one six and the other dice even. For each dice fixed as 6 we have 3 possibilities for the other. Removing the repeated double six this gives us 5 possibilities out of 32 total ones, thus

P(EF) = 5/32

If one dice is 6 and both dices are equal, then we have double six, as a result there is only one combination possible out of 32, therefore

P(FG) = 1/32

If both dices are equal, in particular the sum will be even, this means that G= EG, and as a consecuence

P(EG) = P(G) = 1/6

6 0
3 years ago
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