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Crank
3 years ago
6

PLZ HELP GEOMETRY BELOW

Mathematics
1 answer:
katen-ka-za [31]3 years ago
7 0

This is known as Einstein's proof, not because he was the first to come up with it, but because he came up with it as a 15 year old boy.

Here the problem is justification step 2.  The written equation

BC ÷ DC = BC ÷ AC

is incorrect, and wouldn't get us our statement 2, which is correct.

For similar triangles we have to carefully pair the corresponding parts to get our ratios right:

ABC ~ BDC means AB:BD = BC:DC = AC:BC so BC/DC=AC/BC.

Justification 2 has the final division upside down.


You might be interested in
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
Tamiku [17]

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
Which theorem is used here <br>SAS<br>ASA<br>AAS<br>SAS<br>HL​
dmitriy555 [2]

Answer:

ASA i believe

Step-by-step explanation:

there are two angles that are equal then two sides then vertical angles

7 0
3 years ago
List the sides in order from shortest to longest in ∆ABC, with m angle A = 80, m angleB = 3x + 5, and m angle C = 5x - 1.
galina1969 [7]
M<A + m<B + m<C = 180
80 + 3x + 5 + <span>5x - 1 = 180
8x + 84 = 180
8x =96
x = 12

m<</span><span>B = 3x + 5 = 3(12) + 5 = 41
m<C = </span><span>5x - 1 = 5(12) - 1 =59
and 
m<A = 80

</span><span>answer:

order from shortest to longest in ∆ABC</span><span>
AC, AB, BC

P.S.
Smallest angle, shortest side.....
Largest angle, longer side....

</span>
7 0
4 years ago
Please help! Thank youu
alexandr402 [8]

Answer:

Step-by-step explanation:

(0, 4) (2, 0)

(0-4)/(2-0) = -4/2 = -2

y - 4 = -2(x - 0)

y - 4 = -2x + 0

y = -2x + 4

8 0
3 years ago
Which of the values for x and y make the equation 2x + 3y + 4 = 15 true?
hjlf
X=1, y=3 because 2x1 is 2 and 3x3 is 9, 2 plus 9 equals 11 and 11 plus 4 equals 15
5 0
3 years ago
Read 2 more answers
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