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4 years ago

What is 562949953421312 multiplied by 2

Mathematics

1 answer:

Nesterboy [21]4 years ago

6 0

I used and got 1.1258999e+15 lol Im sorry Im only in middle school

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Find the midpoint of the segment with the following endpoints.<br> (7, -2) (1, -10)

dem82 [27]

Answer:

Step-by-step explanation:

The formula of a midpoint of the segment AB:

$M_{AB}\left(\dfrac{x_A+x_B}{2};\ \dfrac{y_A+y_B}{2}\right)$

We have A(7, -2) and B(1, -10). Substitute:

$M_{AB}\left(\dfrac{7+1}{2};\ \dfrac{-2+(-10)}{2}\right)\\\\M_{AB}\left(\dfrac{8}{2};\ \dfrac{-12}{2}\right)\\\\M_{AB}(4;\ -6)$

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3 years ago

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Distance between (-6, -9) and (-20, -57)

Paul [167]

Answer:

Step-by-step explanation:

(-20,-57)-(-6,-9)=(-26,-66)

-26+-66=-92

|-92|

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4 years ago

To get from your house to your grandparent’s house you must avoid walking through a pond. To avoid the pond, you walk 5 meters s

marishachu [46]

The meters that would be saved is 17 meters

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3 years ago

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What is the LCM and GCF for 15 and 60

slavikrds [6]

LCM of 15 and 60 is 60.

GCF of 15 and 60 is 15.

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3 years ago

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Prove De Morgan's law by showing that each side is a subset of the other side by considering x ∈ A⎯⎯⎯ A ¯ ∩ B⎯⎯⎯ B ¯ .

adelina 88 [10]

Solution :

We have to prove that $\overline{A \cup B} = \overline{A} \cap \overline{B}$ (De-Morgan's law)

Let $x \in \bar{A} \cap \bar{B},$ then $x \in \bar{A}$ and $x \in \bar{B}$

and so $x \notin \bar{A}$ and $x \notin \bar{B}$ .

Thus, $x \notin A \cup B$ and so $x \in \overline{A \cup B}$

Hence, $\bar{A} \cap \bar{B} \subset \overline{A \cup B}$ .........(1)

Now we will show that $\overline{A \cup B} \subset \overline{A} \cap \overline{B}$

Let $x \in \overline{A \cup B}$ ⇒ $x \notin A \cup B$

Thus x is present neither in the set A nor in the set B, so by definition of the union of the sets, by definition of the complement.

$x \in \overline{A}$ and $x \in \overline{B}$

Therefore, $x \in \overline{A} \cap \overline{B}$ and we have $\overline{A \cup B} \subset \overline{A} \cap \overline{B}$ .............(2)

From (1) and (2),

$\overline{A \cup B} = \overline{A} \cap \overline{B}$

Hence proved.

3 0

3 years ago