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lapo4ka [179]
3 years ago
8

What does -5+3x equals

Mathematics
2 answers:
SCORPION-xisa [38]3 years ago
7 0

Answer:

10

Step-by-step explanation:

10-10=0

e-lub [12.9K]3 years ago
3 0

Answer:

the answer is -15

Step-by-step explanation:

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Find the area of an equilateral triangle with a side of 6 inches.
sweet-ann [11.9K]

Answer:

9\sqrt{3} in²

Step-by-step explanation:

The area (A) of an equilateral triangle is calculated as

A = \frac{s^2\sqrt{3} }{4} ( s is the length of a side )

Here s = 6, thus

A = \frac{6^2\sqrt{3} }{4} = \frac{36\sqrt{3} }{4} = 9\sqrt{3} in²

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3 years ago
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I think it would be C

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Prove the following DeMorgan's laws: if LaTeX: XX, LaTeX: AA and LaTeX: BB are sets and LaTeX: \{A_i: i\in I\} {Ai:i∈I} is a fam
MariettaO [177]
  • X-(A\cup B)=(X-A)\cap(X-B)

I'll assume the usual definition of set difference, X-A=\{x\in X,x\not\in A\}.

Let x\in X-(A\cup B). Then x\in X and x\not\in(A\cup B). If x\not\in(A\cup B), then x\not\in A and x\not\in B. This means x\in X,x\not\in A and x\in X,x\not\in B, so it follows that x\in(X-A)\cap(X-B). Hence X-(A\cup B)\subset(X-A)\cap(X-B).

Now let x\in(X-A)\cap(X-B). Then x\in X-A and x\in X-B. By definition of set difference, x\in X,x\not\in A and x\in X,x\not\in B. Since x\not A,x\not\in B, we have x\not\in(A\cup B), and so x\in X-(A\cup B). Hence (X-A)\cap(X-B)\subset X-(A\cup B).

The two sets are subsets of one another, so they must be equal.

  • X-\left(\bigcup\limits_{i\in I}A_i\right)=\bigcap\limits_{i\in I}(X-A_i)

The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices i\in I.

Proof of one direction for example:

Let x\in X-\left(\bigcup\limits_{i\in I}A_i\right). Then x\in X and x\not\in\bigcup\limits_{i\in I}A_i, which in turn means x\not\in A_i for all i\in I. This means x\in X,x\not\in A_{i_1}, and x\in X,x\not\in A_{i_2}, and so on, where \{i_1,i_2,\ldots\}\subset I, for all i\in I. This means x\in X-A_{i_1}, and x\in X-A_{i_2}, and so on, so x\in\bigcap\limits_{i\in I}(X-A_i). Hence X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i).

4 0
3 years ago
Two Trains left a station at the same time. One traveled north at a certain speed and the other traveled south at twice that spe
sashaice [31]
Let s = northbound train
Then
2s = southbound train
:
Distance = time * speed
4s + 4(2s) = 600
:
4s + 8s = 600
:
12s = 600
:
s = 600/12
:
s = 50 mph is the northbound train
Then
2(50) = 100 mph is the southbound train
:
:
Check:
4(50) + 4(100) = 600
3 0
3 years ago
Read 2 more answers
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