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Dominik [7]
3 years ago
6

calculeaza lungimea segmentului ab in fiecare dintre cazuri:A(1,5);B(4,5);A(2,-5),B(2,7);A(3,1)B(-1,4);A(-2,-5)B(3,7);A(5,4);B(-

3,-2);A(1,-8);B(-5,0)

Mathematics
1 answer:
Tatiana [17]3 years ago
6 0

Answer:

1. 3; 2. 12; 3. 5; 4. 13; 5. 10; 6. 10

Step-by-step explanation:

We can use the distance formula to calculate the lengths of the line segments.

d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}

1. A (1,5), B (4,5) (red)

d = \sqrt{(x_{2} - x_{1}^{2}) + (y_{2} - y_{1})^{2}} = \sqrt{(4 - 1)^{2} + (5 - 5)^{2}}\\= \sqrt{3^{2} + 0^{2}} = \sqrt{9 + 0} = \sqrt{9} = \mathbf{3}

2. A (2,-5), B (2,7) (blue)

d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(2 - 2)^{2} + (7 - (-5))^{2}}\\= \sqrt{0^{2} + 12^{2}} = \sqrt{0 + 144} = \sqrt{144} = \mathbf{12}

3. A (3,1), B (-1,4 ) (green)

d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(-1 - 3)^{2} + (4 - 1)^{2}}\\= \sqrt{(-4)^{2} + 3^{2}} = \sqrt{16 + 9} = \sqrt{25} = \mathbf{5}

4. A (-2,-5), B (3,7) (orange)

d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(3 - (-2))^{2} + (7 - (-5))^{2}}\\= \sqrt{5^{2} + 12^{2}} = \sqrt{25 + 144} = \sqrt{169} = \mathbf{13}

5. A (5,4), B (-3,-2) (purple)

d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(-3 - 5)^{2} + (-2 - 4)^{2}}\\= \sqrt{(-8)^{2} + (-6)^{2}} = \sqrt{64 + 36} = \sqrt{100} = \mathbf{10}

6. A (1,-8), B (-5,0) (black)

d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(-5 - 1)^{2} + (0 - (-8))^{2}}\\-= \sqrt{(-6)^{2} + (-8)^{2}} = \sqrt{36 + 64} = \sqrt{100} = \mathbf{10}

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27 + (8-5) -am looking for the answer of number 7
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Answer:

The answer to the equation from question 7 is 14.

Step-by-step explanation:

In question 7, we are given an equation.

2³ + (8 - 5)² - 3

First, subtract 5 from 8 in the parentheses.

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Add 8 to 9.

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

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

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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

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3 years ago
Light travels at 3.0 x 108 km/s. How far does light travel in 5 minutes?
masya89 [10]
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....................................................

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2 years ago
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Westkost [7]

To find the area of the shaded part we have to subtract the rectangle area from the triangle area :

s =  \frac{1}{2}  \times (8x + 2)(6x) - (x)(3x + 1) \\

s =  \frac{1}{2}  \times (48 {x}^{2}  + 12x) - (3 {x}^{2}  + x) \\

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s = 21 {x}^{2}  + 5x

And we're done

Have a great day ♡

4 0
2 years ago
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