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allsm [11]
3 years ago
7

La raíz de un número real negativo existe en los reales si y sólo si su índice es:

Mathematics
1 answer:
-BARSIC- [3]3 years ago
4 0

Respuesta:

C) impar

Explicación paso a paso:

Solo se puede con un índice impar, ya que al ser negativo su solución no existe en los reales si tiene índice par

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inessss [21]
X = 5
Move the variable then cells to like terms and simplify
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3 years ago
What is the product in scientific notation?
sashaice [31]
(2.5 × 10^{-10} ) × (7 × 10^{-6} )

First, simplify brackets. / Your problem should look like: 2.5 × 10^{-10} × 7 × 10^{-6}
Second, simplify. / Your problem should look like: 17.5 × 10^{-10} × 10^{-6}
Third, simplify exponent. / Your problem should look like: 17.5 × 10^{-15}

Answer: B

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3 years ago
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Work out the area of a rectangle with base, b = 20mm and perimeter, P = 60mm.
skelet666 [1.2K]

base+perimeter

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8 0
3 years ago
Como simplifica essa expressão: x+6y-7z+2-8x-12y-10z-20:​
n200080 [17]

Answer:

-17 z - 6 y - 7 x + -18

Step-by-step explanation:

Simplify the following:

x + 6 y - 7 z + 2 - 8 x - 12 y - 10 z - 20

Hint: | Group like terms in x + 6 y - 7 z + 2 - 8 x - 12 y - 10 z - 20.

Grouping like terms, x + 6 y - 7 z + 2 - 8 x - 12 y - 10 z - 20 = (-7 z - 10 z) + (6 y - 12 y) + (x - 8 x) + (2 - 20):

(-7 z - 10 z) + (6 y - 12 y) + (x - 8 x) + (2 - 20)

Hint: | Combine like terms in -7 z - 10 z.

-7 z - 10 z = -17 z:

-17 z + (6 y - 12 y) + (x - 8 x) + (2 - 20)

Hint: | Combine like terms in 6 y - 12 y.

6 y - 12 y = -6 y:

-17 z + -6 y + (x - 8 x) + (2 - 20)

Hint: | Combine like terms in x - 8 x.

x - 8 x = -7 x:

-17 z - 6 y + -7 x + (2 - 20)

Hint: | Evaluate 2 - 20.

2 - 20 = -18:

Answer: -17 z - 6 y - 7 x + -18

3 0
3 years ago
WILL MARK BRAINIEST!!1 PLZ HELP!25 points!!! The coordinate grid shows points A through K. What point is a solution to the syste
marishachu [46]

Answer:

A, C, D and K

Step-by-step explanation:

Inequalities:

y ≤ −2x + 10  and  y >1/2x-2

<u>Finding zeros for the first inequality:</u>

  • x=0 ⇒ y= 10, point (0, 10)
  • y=0 ⇒ x= 5, point (5, 0)
  • Locating these 2 points and connecting to get the first line.

Space to the left of this line (as y≤ ) is the solution for this inequality, any points on this line are inclusive.

<u>Finding zeros for the second inequality:</u>

  • x=0 ⇒ y= -2, point (0, -2)
  • y= 0 ⇒ x= 4, point (4, 0)
  • Locating these points and connecting to get the second line.

Space above this line (as y > sign) is the solution for this inequality, any points on this line are not inclusive.

Now we have the space of intersection of the above inequalities.

So the points in same section are the solution.

They are:

  • Points A, C, D and K

See attached for explanation

7 0
4 years ago
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