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Salsk061 [2.6K]
3 years ago
6

En el experimento Extraer una carta de una baraja española se consideran estos sucesos: A= 《Salir as 》 B= 《salir bastos 》 Calcul

a A U B y A interseccion B. ¿Son compatibles A y B ?
Mathematics
1 answer:
Ganezh [65]3 years ago
7 0

Answer:

They are compatible

Step-by-step explanation:

The first thing is to say that an "ace" and that it is a "coarse"

"ace" is card number 1. Group A

"coarse" is a type of the deck, found from number 1 to card 13. Group B

Thus:

 Calculate A U B:

1 to 13 + 1 of the other types of cards in the deck.

At intersection B:

1 of "coarse"

Therefore, if group A is compatible with group B

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Solve the equation 2p/3-12=-2
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Hey there! 

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Firstly, we are going add 12 on each of the sides that we're working with. like ↓ 

p - 12 + 12 \\ \\ = -2 + 12

This gives us \frac{2}{3}p = 10 (if you are wondering how we got the out come of 10 it is because I added -2 + 12 = 10

Now multiply \frac{3}{2} on each of your sides 

\frac{3}{2} ( \frac{2}{3}p)  \\ \\  =  \frac{3}{2} (10)

Cancel the first set and you will find the value of p

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Robin is flying a kite. She ties the 50 foot kite string to the ground. If the kite is
krok68 [10]

\bold{\huge{\orange{\underline{ Solution }}}}

<h3><u>Given </u><u>:</u><u>-</u></h3>

  • <u>We </u><u>have </u><u>given </u><u>in </u><u>the </u><u>question </u><u>that</u><u>, </u><u> </u><u>Robin </u><u>is </u><u>flying </u><u>a </u><u>kite</u><u>. </u>
  • <u>She </u><u>ties </u><u>the </u><u>5</u><u>0</u><u> </u><u>foot </u><u>kite </u><u>string </u><u>to </u><u>the </u><u>ground </u><u>.</u>
  • <u>The</u><u> </u><u>kite </u><u>is </u><u>flying </u><u>at </u><u>4</u><u>0</u><u> </u><u>feet </u><u>high </u>

<h3><u>To </u><u>Find </u><u>:</u><u>-</u></h3>

  • <u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>measure </u><u>of </u><u>the </u><u>angle </u><u>the </u><u>string </u><u>forms </u><u>with </u><u>the </u><u>group </u><u>that </u><u>is </u><u>angle </u><u>of </u><u>elevation</u><u>. </u>

<h3><u>Let's </u><u>Begin </u><u>:</u><u>-</u></h3>

<u>According </u><u>to </u><u>the </u><u>given</u><u> </u><u>question</u><u>, </u>

  • Hypotenuse AC ( Distance of string from the ground) = 50 ft.
  • Perpendicular height AB ( Distance of the kite from the ground) = 40 ft.

<h3><u>Therefore</u><u>, </u></h3>

<u>By </u><u>using </u><u>trigonometric </u><u>ratios</u><u>,</u><u> </u>

{ \bold{\pink{ Sin Φ = }}{\bold{\pink{\dfrac{Perpendicular}{Hypotenuse }}}}

\bold{\red{ Cos Φ = }}{\bold{\red{\dfrac{Base}{Hypotenuse }}}}

\bold{\purple{ Sin Φ = }}{\bold{\purple{\dfrac{Perpendicular}{Base }}}} }

<u>The </u><u>Angle </u><u>of </u><u>elevation </u><u>will </u><u>be </u>

<u>[</u><u> </u><u>The </u><u>angle </u><u>that </u><u>is </u><u>formed </u><u>between </u><u>the </u><u>line </u><u>of </u><u>sight </u><u>and </u><u>base </u><u>of </u><u>the </u><u>triangle </u><u>is </u><u>called </u><u>angle </u><u>of </u><u>elevation </u><u>]</u>

\bold{\pink{Sin Φ = }}{\bold{\pink{\dfrac{AB}{AC }}}}

\sf{ Sin Φ = }{\sf{\dfrac{40}{50 }}}

\sf{ Sin Φ = }{\sf{\dfrac{4}{5}}}

\sf{ Sin Φ = 0.8 }

\bold{\green{  Φ = 45.83° }}

Hence, The measure of the angle the string forms with the ground is 45.83° .

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