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hjlf
3 years ago
7

ABCD is a parallelogram and this figure is NOT drawn to scale.

Mathematics
2 answers:
dimulka [17.4K]3 years ago
7 0
The angle does not matter. Think of it as finding the other side to a triangle. Use a ^{2} +b^2 =c^2

a=39 (line AB)
b=b (the leg we need to find)
c=89 (line BD)

39^2 + b^2 = 89^2
1521 + b^2 = 7921
(subtract the 1521 from both sides)
b^2 = 6400
(square root both sides)
\sqrt{(b^2)} =  \sqrt{6400} 


b = 80
AD=80
mezya [45]3 years ago
7 0

Answer:  The length of AD is 80 cm.

Step-by-step explanation:  Given that ABCD is a parallelogram, where m∠BCD = 90°, AB = 39 cm and BD = 89 cm.

We are to find the value of AD.

We know that

the measures of the opposite angles of a parallelogram are equal.

So, in the parallelogram ABCD, we have

m∠BAD =  m∠BCD = 90°.

So, triangle ABD is a right-angled triangle  with AB and AD as two legs and BD as the hypotenuse.

Using Pythagoras law in triangle ABD, we get

AB^2+AD^2=BD^2\\\\\Rightarrow AD^2=BD^2-AB^2\\\\\Rightarrow AD^2=89^2-39^2\\\\\Rightarrow AD^2=7921-1521\\\\\Rightarrow AD^2=6400\\\\\Rightarrow AD^2=80^2\\\\\Rightarrow AD=80.

Thus, the length of AD is 80 cm.

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204,000 grapefruit.

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10x20,400=204,000

Don't forget the units! In this problem, the units are grapefruit.
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What is 7,776 ÷ 24? This is for 30 points.
baherus [9]

Answer:

324

Step-by-step explanation:

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julia-pushkina [17]

Hello there.

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Remove parenthesis.

5-34

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Hope this helps!

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In a football or soccer game, you have 22 players, from both teams, in the field. what is the probability of having at least any
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We can solve this problem using complementary events. Two events are said to be complementary if one negates the other, i.e. E and F are complementary if

E \cap F = \emptyset,\quad E \cup F = \Omega

where \Omega is the whole sample space.

This implies that

P(E) + P(F) = P(\Omega)=1 \implies P(E) = 1-P(F)

So, let's compute the probability that all 22 footballer were born on different days.

The first footballer can be born on any day, since we have no restrictions so far. Since we're using numbers from 1 to 365 to represent days, let's say that the first footballer was born on the day d_1.

The second footballer can be born on any other day, so he has 364 possible birthdays:

d_2 \in \{1,2,3,\ldots 365\} \setminus \{d_1\}

the probability for the first two footballers to be born on two different days is thus

1 \cdot \dfrac{364}{365} = \dfrac{364}{365}

Similarly, the third footballer can be born on any day, except d_1 and d_2:

d_3 \in \{1,2,3,\ldots 365\} \setminus \{d_1,d_2\}

so, the probability for the first three footballers to be born on three different days is

1 \cdot \dfrac{364}{365} \cdot \dfrac{363}{365}

And so on. With each new footballer we include, we have less and less options out of the 365 days, since more and more days will be already occupied by another footballer, and we can't have two players born on the same day.

The probability of all 22 footballers being born on 22 different days is thus

\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}

So, the probability that at least two footballers are born on the same day is

1-\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}

since the two events are complementary.

8 0
3 years ago
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