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Olin [163]
3 years ago
5

Simplify 12p-8+6p+14

Mathematics
2 answers:
Klio2033 [76]3 years ago
5 0
18p+6 would be your answer
12p+6p=18p
-8+14=6
Tatiana [17]3 years ago
3 0
Combine the like terms: 12p & 6p ; 8 & 14

12p - 6p = 6p

8 + 14 = 22 ; -8 + 14 = 6

Problem becomes and your answer is ➡
18z + 6
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Which statement is true <br> -11&gt;-10<br> 2&lt;-5<br> -4&gt;-10<br> -6&lt;-1
Masja [62]

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the true statement is -4>-10 and -6<-1

Step-by-step explanation:

it's because in negative numbers the greater the number goes the lower the actual value is

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2 years ago
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Prove the following statements: (a) For every integer x, if x is even, then for every integer y, xy is even. (b) For every integ
Murrr4er [49]

Answer and Step-by-step explanation:

(a) Given that x and y is even, we want to prove that xy is also even.

For x and y to be even, x and y have to be a multiple of 2. Let x = 2k and y = 2p where k and p are real numbers. xy = 2k x 2p = 4kp = 2(2kp). The product is a multiple of 2, this means the number is also even. Hence xy is even when x and y are even.

(b) in reality, if an odd number multiplies and odd number, the result is also an odd number. Therefore, the question is wrong. I assume they wanted to ask for the proof that the product is also odd. If that's the case, then this is the proof:

Given that x and y are odd, we want to prove that xy is odd. For x and y to be odd, they have to be multiples of 2 with 1 added to it. Therefore, suppose x = 2k + 1 and y = 2p + 1 then xy = (2k + 1)(2p + 1) = 4kp + 2k + 2p + 1 = 2(kp + k + p) + 1. Let kp + k + p = q, then we have 2q + 1 which is also odd.

(c) Given that x is odd we want to prove that 3x is also odd. Firstly, we've proven above that xy is odd if x and y are odd. 3 is an odd number and we are told that x is odd. Therefore it follows from the second proof that 3x is also odd.

3 0
3 years ago
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