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OverLord2011
2 months ago
10

Paul is 2 years younger than patricia. Daniel is 25% older than Patricia. Ten years ago, Daniel was 50% older than Patricia. How

old is paul currently.
Mathematics
1 answer:
Leona [12.6K]2 months ago
7 0

Daniel is 25% older than Patricia. Ten years prior, Daniel was 50% older than Patricia.

This results in the equations: 1.25p = d and 1.5(p - 10) = d - 10. Solving these yields Daniel's current age as 25 and Patricia's as 20.

Since Paul is two years younger than Patricia:

20 - 2 = 18



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(4 – 2i)(6 + 2i)

Use FOIL

First: 4 multiplied by 6 equals 24

Outer: 4 times 2i gives 8i

Inner: -2i times 6 results in -12i

Last: -2i times 2i equals -4i², which simplifies to 4

Combine the results

24 + 8i - 12i + 4

Final result: 28 - 4i

The answer is: 28 - 4i

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Describe the sample space for each of these experiments (a) A coin is tossed and a single die is rolled. (b) A paper cup is toss
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Answer:

a) S={(head,1),(head,2),(head,3),(head,4),(head,5),(head,6),(tails,1),(tails,2),(tails,3),(tails,4),(tails,5),(tails,6)}

b) S={top,bottom,side}

c) S={(Ace of diamonds),(Two of diamonds),(Three of diamonds),(Four of diamonds),(Five of diamonds),(Six of diamonds),(Seven of diamonds), (Eight of diamonds),(Nine of diamonds),(Ten of diamonds), (Jack of diamonds), (Queen of diamonds),(King of diamonds),(Ace of hearts),(Two of hearts),(Three of hearts),(Four of hearts),(Five of hearts),(Six of hearts),(Seven of hearts), (Eight of hearts), (Nine of hearts),(Ten of hearts), (Jack of hearts), (Queen of hearts),(King of hearts),(Ace of clubs),(Two of clubs),(Three of clubs),(Four of clubs),(Five of clubs),(Six of clubs),(Seven of clubs), (Eight of clubs), (Nine of clubs),(Ten of clubs), (Jack of clubs),(Queen of clubs),(King of clubs),(Ace of spades),(Two of spades),(Three of spades),(Four of spades),(Five of spades),(Six of spades),(Seven of spades),(Eight of spades), (Nine of spades),(Ten of spades),(Jack of spades),(Queen of spades),(King of spades)}

Step-by-step explanation:

The sample space consists of all potential outcomes for each experiment:

a) The sample encompasses the outcomes resulting from tossing a coin (heads or tails) and throwing a die (a number between 1 and 6):

S={(head,1),(head,2),(head,3),(head,4),(head,5),(head,6),(tails,1),(tails,2),(tails,3),(tails,4),(tails,5),(tails,6)}

b) A paper cup can settle in one of three different positions: on its top, bottom, or side:

S={top,bottom,side}

c) The entire set of cards within a standard deck constitutes the sample space:

S={(Ace of diamonds),(Two of diamonds),(Three of diamonds),(Four of diamonds),(Five of diamonds),(Six of diamonds),(Seven of diamonds), (Eight of diamonds),(Nine of diamonds),(Ten of diamonds), (Jack of diamonds), (Queen of diamonds),(King of diamonds),(Ace of hearts),(Two of hearts),(Three of hearts),(Four of hearts),(Five of hearts),(Six of hearts),(Seven of hearts), (Eight of hearts), (Nine of hearts),(Ten of hearts), (Jack of hearts), (Queen of hearts),(King of hearts),(Ace of clubs),(Two of clubs),(Three of clubs),(Four of clubs),(Five of clubs),(Six of clubs),(Seven of clubs), (Eight of clubs), (Nine of clubs),(Ten of clubs), (Jack of clubs),(Queen of clubs),(King of clubs),(Ace of spades),(Two of spades),(Three of spades),(Four of spades),(Five of spades),(Six of spades),(Seven of spades),(Eight of spades), (Nine of spades),(Ten of spades),(Jack of spades),(Queen of spades),(King of spades)}

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Important details about isosceles triangle ABC:

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  • In isosceles triangle ABC, the sides AB and BC are equal, meaning AC=BC.
  • The base angles at AB are equal, m∠A=m∠B=30°.

1. Consider the right triangle ACD. The angle adjacent to side AD is 30°, which dictates that the hypotenuse AC is double the length of the opposite side CD relating to angle A.

AC=2CD.

2. Now, for right triangle BCD, the angle next to side BD is also 30°, so hypotenuse BC is twice the opposite leg CD linked to angle B.

BC=2CD.

3. To calculate the perimeters of triangles ACD, BCD, and ABC:

P_{ACD}=AC+CD+AD=2CD+CD+AD=3CD+AD;

P_{BCD}=BC+CD+BD=2CD+CD+AD=3CD+AD;

P_{ABC}=AC+BC+AB=2CD+2CD+AD+BD=4CD+2AD.

4. If the total of the perimeters of triangles ACD and BCD is 20 cm greater than the perimeter of triangle ABC, then

P_{ACD}+P_{BCD}=P_{ABC}+20,\\ \\3CD+AD+3CD+AD=4CD+2AD+20,\\ \\6CD+2AD=4CD+2AD+20,\\ \\2CD=20.

5. Given that AC=BC=2CD, the lengths of legs AC and BC of the isosceles triangles are 20 cm.

Answer: 20 cm.

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