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2 years ago

Use the value of the discriminant to determine the type of roots for x²-3x+1=0

Mathematics

1 answer:

nignag [31]2 years ago

7 0

Hi,

Δ = 9 - 4 * 1 * 1 = 5

the roots are :

x1 = (3 - √5) / 2

x2 = (3 + √5) / 2

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Calculate (A⃗ ×B⃗ )⋅C⃗ for the three vectors A⃗ with magnitude A = 5.08 and angle θA = 25.6 ∘ measured in the sense from the +x

alexgriva [62]

Answer:

(A⃗ ×B⃗ )⋅C⃗ = - 76.415

Step-by-step explanation:

First we need to calculate (A⃗ ×B⃗ ) :

(A⃗ ×B⃗ ) = A.B.sin (α).n

Where A is the magnitude of A⃗

Where B is the magnitude of B⃗

Where α is the angle between A⃗ and B⃗ = 63.9 - 25.6 = 38.3

Finally n is the vector orthogonal to A⃗ and B⃗

n magnitude is 1 and his direction is given by the right hand-rule

so n = ( 0 , 0 , 1 )

(A⃗ ×B⃗ ) = A.B.sin (α).n = 5.08 . 3.94 . sin (38.3) . (0 , 0 , 1 ) = (0,0,12.4)

C⃗ can be written as C.(0,0,-1) because of his +z - direction

C.(0,0,-1) = 6.16.(0,0,-1) = (0,0,-6.16)

(A⃗ ×B⃗ )⋅C⃗ = (0,0,12.4).(0,0,-6.16) = -76.41480787 = -76.415

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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

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

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