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Aleks [24]
3 years ago
8

Expand and simplify 4 (x-3)-2 (1-x)

Mathematics
2 answers:
oee [108]3 years ago
6 0
2(x + 5) 4(x + 3) - 2(x + 2) -2(2x - 3) 7(3 - 2x) 7x + 29 10x - 5 -6x - 15 12 + 3x 2(2x + 7) -2(3 + 4x) 4(x - 3) - 2(x + 2) 5(3 - x) 8x + 4 2x - 16 7x - 7 7x + 21 4x + 3(2x - 1) 3(4 + x) 5(x + 5) - 2(x + 2) 4(x + 3) - 2(x - 2) 2x + 16 2x + 10 2x - 10 -4x + 6 -3(2x + 5) 4(2x + 1) 7(x - 1) -7(3 - 2x) -6 - 8x 10x - 3 4x + 14 2x + 8 5(x + 5) + 2(x + 2) 4(x - 3) - 2(x - 1) 5(2x - 1) 5(x + 5) + 2(x - 2) 15 - 5x 3x + 21 -21 + 14x 21 - 14x
Nikolay [14]3 years ago
4 0
<span> </span><span>4(x-3)-2(1-x) 
distribute 
4x -12 -2 + x 
combine 
5x - 14</span>
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What is the solution to the inequality lx-4l=3
xxMikexx [17]

Answer: x = 7, x = 1

Step-by-step explanation:

|x-4| = 3

we know either x-4 = 3 or x-4 = -3

x-4 + 4 = 3 + 4

x = 7

x - 4 + 4 = -3 + 4

x = 1

8 0
1 year ago
Can you help me please?
gulaghasi [49]

Answer:

I think the answer is B, because the value does increase by one

6 0
3 years ago
Jamie says that the expression 6x-2x+4 and 4(x+1) are not equivalent because one expression has a term that is subtracted and th
Sati [7]

Answer:

1. D 2. A 3. Distribute the four in 4(x+1)

Step-by-step explanation:

1. It says that Jamie is saying that they are "not equivalent".

2. You have to use the distributive property to distribute the 4 to x and to the 1.

3. After distributing, compare the two expressions and now see if they are equivalent or not.

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3 years ago
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lesya692 [45]
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3 0
3 years ago
Read 2 more answers
A computer system uses passwords that are six characters, and each character is one of the 26 letters (a–z) or 10 integers (0–9)
Blababa [14]

First of all, since we have 36 characters available per spot (26 letters and 10 digits), and we have 6 spots, we have a total of

36^6

possible passwords.

Event A happens if the password starts with either a, e, i, o or u. If we fix the first character, we're left with 36 characters available for each of the remaining 5 spots, leading to a total of

5\cdot 36^5

possible passwords.

So, the probability of event A, computed as the ratio between "good" cases and all possible cases, is

\dfrac{5\cdot 36^5}{36^6}=\dfrac{5}{36}

Event B works exactly the same, since we're fixing the last spot, leaving 36 characters available for each of the first 5 spots. So, we have

P(A)=P(B)=\dfrac{5}{36}

As for the intersection, we want the first character to be a vowel, and the last character to be an even digits. There are 25 passwords satisfying this request:

axxxx0,\ axxxx2,\ axxxx4,\ axxxx6,\ axxxx8

exxxx0,\ exxxx2,\ exxxx4,\ exxxx6,\ exxxx8

ixxxx0,\ ixxxx2,\ ixxxx4,\ ixxxx6,\ ixxxx8

oaxxxx0,\ oxxxx2,\ oxxxx4,\ oxxxx6,\ oxxxx8

uxxxx0,\ uxxxx2,\ uxxxx4,\ uxxxx6,\ uxxxx8

Where x can be any of the 36 characters.

So, we have 25 cases with 4 vacant slots, leading to a probability of

P(A\cap B)=\dfrac{25\cdot 36^4}{36^6}=\dfrac{25}{1296}

Finally, you can compute the probability of the union using the formula

P(A\cup B)=P(A)+P(B)-P(A\cap B)

Since we already computed all these quantities.

7 0
3 years ago
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