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irina1246 [14]
3 years ago
11

Please could someone please explain how i do this, much appreciated!

Mathematics
1 answer:
Vilka [71]3 years ago
6 0
You want to multiply together the number of possible choices for each digit in the code.

0-9 is ten possible numbers, since repetitions are allowed (example code 2,2,2,2) there are ten possible choices per digit in the code.

10*10*10*10 = 10,000 four digit coded possible
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Find the slope of the line in the image.<br> someone plz helpppp
pogonyaev

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Step-by-step explanation:(-2,6)(2,4)

M=y2-y1

   x2-x1

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2-(-2)   =  4

The slope is -2/4

4 0
3 years ago
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Apply the distributive property to factor out the greatest common factor. 12+80=12+80
Ludmilka [50]

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Step-by-step explanation:

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Given a function I and a subset A of its domain, let I(A) represent the range of lover the set A; that is, I(A) = {I(x) : x E A}
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Step-by-step explanation:

(a)

Using the definition given from the problem

f(A) = \{x^2  \, : \, x \in [0,2]\} = [0,4]\\f(B) = \{x^2  \, : \, x \in [1,4]\} = [1,16]\\f(A) \cap f(B) = [1,4]  = f(A \cap B)\\

Therefore it is true for intersection. Now for union, we have that

A \cup B = [0,4]\\f(A\cup B ) = [0,16]\\f(A) = [0,4]\\f(B)= [1,16]\\f(A) \cup f(B) = [0,16]

Therefore, for this case, it would be true that f(A\cup B) = f(A)\cup f(B).

(b)

1 is not a set.

(c)

To begin with  

A\cap B \subset A,B

Therefore

g(A\cap B) \subset g(A) \cap g(B)

Now, given an element of g(A) \cap g(B) it will belong to both sets, therefore it also belongs to g(A\cap B), and you would have that

g(A)\cap g(B) \subset  g(A \cap B), therefore  g(A)\cap g(B)  =  g(A \cap B).

(d)

To begin with A,B  \subset A \cup B, therefore

g(A) \cup g(b) \subset g(A\cup B)

7 0
3 years ago
Brooke found the equation of the line passing through the points (–7, 25) and (–4, 13) in slope-intercept form as follows. Step
Katen [24]

Answer: She mixed up the slope and y-intercept when she wrote the equation in step 3.

Step-by- explanation:

Hope this helps :)

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HELP PLEASE!! WILL CROWN BRAINLIEST.....
Ivenika [448]

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