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olga2289 [7]
3 years ago
7

How do I solve this question (19a)?

Mathematics
1 answer:
fredd [130]3 years ago
8 0
Answer i think would be
l-4x+4l
---------
x-1
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egoroff_w [7]
Never heard of punch white oj
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Diana ,and Jennifer and Christina are making class for the craft fair Diana has made three tenths of her Jennifer has made Two f
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Explain what needs to be fixed.<br> 2n + 10 = -2<br> 2n = 12 Step one<br> n = 6 Step two
Brilliant_brown [7]

Answer:

step one needs to be fixed... you are supposed to subtract 10 from both sides of the equation.

Step-by-step explanation:

STEP 1: Move all terms not containing n to the right side of the equation.

2n=-12

STEP 2: Divide each term by 2 and simplify.

n=-6

8 0
3 years ago
Solve the equation, 3x – 6 = 12, for x. Which of the following properties did you apply? Question 20 options: A) Distributive pr
Luda [366]

Answer:

C) Addition property of equality and division property of equality

Step-by-step explanation:

To find the answer, let's first solve the equation for x...

3x-6=12

Add 6 to both sides (Addition property of equality)

3x=6

Divide both sides by 3 (Division property of equality)

x=2

We used both the addition property of equality and the division property of equality therefore the answer is C

5 0
2 years ago
We select n + 1 different integers from the set { 1 , 2 , ··· , 2 n } . Provethat there will alwaysbe two among the selected inte
just olya [345]

Answer:

See answer below

Step-by-step explanation:

From the set

{1,2,3,4...2n} we have 2n numbers in total , n are odd and n are even , therefore for a sample of n+1 numbers , we have at least 1 even number and 1 odd number.

Then

it the set includes 1 , the largest common divisor is 1 for 1 and the other numbers

if the set includes 3, there will be always a number that is not divisible by 3. Even we construct a set of n+1 numbers that are multiple of 3 , the largest number would be 3*(n+1)= 3*n+3 > 2*n (out of bounds) , therefore we are forced to take other number that is not divisible by 3  → the largest common divisor of that number with 3 is 1

If the set includes any other prime number → the largest common divisor of that with any other is 1

For the remaining odd numbers N, they can be factorised into other 2 odd common divisors N₂ and n₂ :

N = N₂*n₂ , since n₂ ≥ 2 →  N₂ < N

then the even N₂ also should be contained in the set

therefore also for N₂

N = N₃*n₃ →  N₃ < N₂

therefore if we continue , we would obtain a number  even Nn that has no smaller common divisors → since we cannot take all the multiples of N min ( because Nmin*(n+1)= Nmin*n+Nmin > 2*n for Nmin≥2) → there is at least a number in the sample of n+1 integers whose largest common divisor is 1

6 0
3 years ago
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