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Natali [406]
3 years ago
8

Given n=−12.

Mathematics
1 answer:
tamaranim1 [39]3 years ago
8 0

Answer:

A) −31+2n=−55

Step-by-step explanation:

2n−31=−55

2n−31+31=−55+31

2n=−24

2n over 2=-24 over 2

n=-12

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Alik [6]

Answer: 0.5, 48,36,1.5,2.25

Step-by-step explanation:

Dado que, la última fila dice 12: 3, eso significa que la relación entre los dos números es 4: 1, por lo que si aplicamos al resto, por ejemplo, 9 sería 2.25 ya que divides 9 entre 4.

¡Espero que tenga sentido y buena suerte!

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Answer is p y = 1200x + 300
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3 years ago
Given the following relations on the set of all integers where (x,y)∈R if and only if the following is satisfied. (Check ALL cor
Wewaii [24]

Answer:

a) symmetric

b) symmetric, reflexive, transitive

c) antisymmetric

d) symmetric

Step-by-step explanation:

(a) x+y = 0

- this relation is not reflexive, because the only element that relates with itself is 0. x+x = 2x, x+x = 0 only if 2x = 0, hence x = 0. Since 0 relates with itself, then the relation isnt irreflexive either.

Note that for x,y such that x R y, we have that x+y = 0, therefore, y = -x.

-If x,y,z are such that x R y, y R z, then y = -x, z = -y = -(-x) = x. In general x does not relate with z because z=x and the relation isnt reflexive, thus the relation is not transitive. For example, if x = z = 2, y = -2, we have that xRy, yRz, but x does not relate with z.

- This realtion is symmetric due to the commutativity of the sum. If xRy, then  x+y = 0, and y+x = x+y = 0, hence yRx. Therefore, the relation cant be antisymmetric, because every element different from 0 relates to its opposite. For example 2R-2, -2R2 but 2 ≠ -2.

b) x-y is an integer

Since we are taking the substraction of two integers, the result will always be integer. Hence, every pair of elements relate within each other. As a result, the relation is symmetric, reflexive and transitive. However, it is not irreflexive nor antisymmetric, because for example 4R4, and 4R8, 8R4, but 4 is not 8.

c) x = 2y

Note that x = 2x only if x = 0, so the relation is neither reflexive nor irreflexive.

The relation is not symmetric, for example, 4R2 because 4 = 2*2, but 2 does not relate with 4, because 4*2 = 8. However, the relation is antisymmetric, because if xRy, yR2, we have

  • x = 2y
  • y = 2x = 2(2y) = 4y

since y = 4y, y should be 0, and x = 2*0 = 0. Therefore x=y = 0. The relation is antisymmetric.

The relation isnt transitive: 2R4, 4R8, but 2 does not relate with 8 because 8*2 = 16.

d) xy>1

since 0² = 0, 0 does not relate with itself, hence the relation is not reflexive. It is not irreflexive either, because, for example, 2*2 = 4 > 2, thus 2 relates with itself.

The relation is not transitive: 1 relates with every integer greater than itself, but it does not relate with itself, for example 1R7 and 7R1 because 1*7=7*1 = 7 > 1, but 1*1 = 1, it is not greater than 1, hence 1 doesnt relate with itself. This also shows that the relation is not antisymmetric either, because 1R7, 7R1 but 1≠7. The relation, however, is symmetric due to the commutativity of the product. If xy > 1, then yx = xy >1 as well.

I hope that works for you!

8 0
4 years ago
Can anyone help me with this, please?
zloy xaker [14]

9514 1404 393

Answer:

  C.  3x∛(y²z)

Step-by-step explanation:

The relevant rules of exponents are ...

  \sqrt[n]{a^m} = a^\frac{m}{n}\\\\(a^b)^c=a^{bc}

__

Your expression can be rewritten and simplified as follows.

  \displaystyle\sqrt[3]{27x^3y^2z}=(3^3x^3y^2z)^\frac{1}{3}=3xy^\frac{2}{3}z^\frac{1}{3}=3x(y^2z)^\frac{1}{3}=\boxed{3x\sqrt[3]{y^2z}}

6 0
3 years ago
Read 2 more answers
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