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lawyer [7]
3 years ago
13

Order the steps to solve the equation log(x2 − 15) = log(2x) form 1 to 5. x2 − 2x − 15 = 0 Potential solutions are −3 and 5 x2 −

15 = 2x x − 5 = 0 or x + 3 = 0 (x − 5)(x + 3) = 0
Mathematics
2 answers:
Anton [14]3 years ago
7 0
From First to last box.
2
5
1
4
3

aleksley [76]3 years ago
4 0
<h2>Order is 3,1,5,2 and 4.</h2>

Step-by-step explanation:

We have

         log(x² - 15) = log(2x)

Solving the equation

             x² - 15 = 2x

             x² - 2x - 15 = 0

             (x-5)(x+3) = 0

             x -5 = 0   or   x + 3  = 0

              x = 5   or  x = -3

The solutions are −3 and 5.

Order of solution is x² − 15 = 2x, x² - 2x - 15 = 0, (x − 5)(x + 3) = 0, x − 5 = 0 or x + 3 = 0 and potential solutions are −3 and 5.

Order is 3,1,5,2 and 4.

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Joaquin fue al mercado y compro 2,2 kg de papa a S/.1,2 soles el kilo y 3,2 kg de arroz a S/.3,7 soles el kilo, si pagó con un b
nlexa [21]

Responder:

5,52

Explicación paso a paso:

Dado lo siguiente:

Precio por kilo de papa = 1.2

Kg de papa comprada = 2.2kg

Precio por kilo de arroz = 3.7 por kilo

Kg de arroz comprado = 3.2 kg

Valor del boleto emitido para el pago = 20

Precio total de la papa:

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1.2 * 2.2 = 2.64

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Precio * kg comprado

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A bridge hand is made up of 13 cards from a deck of 52. find the probability that a hand chosen at random contains at least 3 ki
Vlad [161]
1.
In total there are C(52, 13) ways that we can pick a hand, that is \frac{52!}{13!39!}


2.
P(a hand contains at least 3 kings)
                =P(a hand contains exactly 3 kings)+P(a hand contains 4 kings)

3.
first let's find P(a hand contains exactly 3 kings):

P(a hand contains exactly 3 kings)
              =n(a hand contains exactly 3 kings)/C(52, 13)
              
n(a hand contains exactly 3 kings)=C(4, 3)*C(48, 10)

where C(4,3) is the total number of ways we can pick 3 out of 4 kings,

C(48, 10) is the number of picking 10 letters to complete a hand, out of the 52-4=48 non-king cards.

so P(a hand contains exactly 3 kings)=[C(4, 3)*C(48, 10)]/C(52, 13)

4. with the same reasoning as in step 3:

P(a hand contains 4 kings)=n(a hand contains 4 kings)/C(52, 13)

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5. 

P(a hand contains at least 3 kings)
                =P(a hand contains exactly 3 kings)+P(a hand contains 4 kings)

=[C(4, 3)*C(48, 10)]/C(52, 13)+ [C(4, 4)*C(48, 9)]/C(52, 13)

=\frac{C(4, 3)*C(48, 10)+C(4, 4)*C(48, 9)}{C(52, 13)}

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simplify by 38! in the denominators and 48! in the numerators :

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now simplify by 9! in all denominators:

\frac{ \frac{4}{10}+  \frac{1}{39} }{ \frac{52*51*50*49}{13*12*11*10*39}}


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