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lakkis [162]
3 years ago
13

Quadrilateral ABCD has coordinates A (3, -5), B (5, -2), C (10, -4), D (8, -7). Quadrilateral ABCD is a

Mathematics
1 answer:
Oksana_A [137]3 years ago
6 0

Answer:

Step-by-step explanation:

Quadrilateral ABCD is parallelogram, because opposite sides are congruent and adjacent sides are not perpendicular.

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A set of equations is given below: Equation C: y = 6x + 9 Equation D: y = 6x + 2 Which of the following best describes the numbe
7nadin3 [17]
So we can get from these two equations that y = 6x + 2 = 6x + 9

6x + 2 = 6x + 9
0 = 7 , but that's incorrect, so this set has no solution.
7 0
3 years ago
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Evaluate the expression 15-n when n =8
Dvinal [7]

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7

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if : n = 2    : 15 -  n = 15 - 8 = 7

4 0
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What is the probability of getting a tails, vowel and an even number?
Zinaida [17]

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\frac{1}{15}

This is assuming the die is from 1 - 5

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Vowel: \frac{1}{3}

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\frac{1}{2}* \frac{1}{3} *\frac{2}{5} =\frac{2}{30}=\frac{1}{15}

8 0
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What is the value of (2.4 x 10^3)+(5.7×10^2)​
nordsb [41]

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4 0
3 years ago
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A) how many ways are there to choose 10 players to take the field?
Kruka [31]

A. C(13, 10) = 13! = 13·12·11 = 13 · 2 · 11 = 286.C(13, 10) = 13! = 13·12·11 = 13 · 2 · 11 = 286.

B. P(13,10)= 13! =13! =13·12·11·10·9·8·7·6·5·4. (13−10)! 3!

C. f there is exactly one woman chosen, this is possible in C(10, 9)C(3, 1) =

10! 3!

9!1! 1!2! 10! 3!

8!2! 2!1! 10! 3!

7!3! 3!0!

= 10 · 3 = 30 ways; two women chosen — in C(10,8)C(3,2) =

= 45·3 = 135 ways; three women chosen — in C(10, 7)C(3, 3) =

= 10·9·8 ·1 = 120 ways. Altogether there are 30+135+120 = 285 1·2·3

<span>possible choices.</span><span> 
</span>

  


5 0
3 years ago
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