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3 years ago

Find the sum (2x−6)+4(x−3) =

Mathematics

2 answers:

xeze [42]3 years ago

7 0

Answer: x=-2

Step-by-step explanation: First we can multiply 4 times anything in the parantheses to get rid of it.

(2x-6)+4(x-3)

(2x-6)+4x-12

Then we combine like terms.

(2x-6)+4x-12

6x-6-12

6x-18

Finally we divide both sides 6.

6x-18

6x/6-18/6

x=-2

The final answer is x=-2

Hope this helps!

Btw this is a backup account

dedylja [7]3 years ago

7 0

Answer:

6x - 18

Step-by-step explanation:

Distribute: 2x − 6 + 4x − 12

Group Common Factors: 2x + 4x - 6 -12

Solve: 6x - 18

I hope this helped, please mark Brainliest, thank you!

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3 years ago

On which number line are -6 and its opposite shown?<br>

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4 years ago

Read 2 more answers

Use the approach in Gauss's Problem to find the following sums of arithmetic

Agata [3.3K]

a. Let S be the first sum,

S = 1 + 2 + 3 + … + 97 + 98 + 99

If we reverse the order of terms, the value of the sum is unchanged:

S = 99 + 98 + 97 + … + 3 + 2 + 1

If we add up the terms in both version of S in the same positions, we end up adding 99 copies of quantities that sum to 100 :

S + S = (1 + 99) + (2 + 98) + … + (98 + 2) + (99 + 1)

2S = 100 + 100 + … + 100 + 100

2S = 99 × 100

S = (99 × 100)/2

Then S has a value of

S = 99 × 50

S = 4950

Aside: Suppose we had n terms in the sum, where n is some arbitrary positive integer. Call this sum ∑(n) (capital sigma). If ∑ is a sum of n terms, and we do the same manipulation as above, we would end up with

2 ∑(n) = n × (n + 1) ⇒ ∑(n) = n (n + 1)/2

b. Let S' be the second sum. It looks a lot like S, but the even numbers are missing. Let's put them back, but also include their negatives so the value of S' is unchanged. In doing so, we have

S' = 1 + 3 + 5 + … + 1001

S' = (1 + 2 + 3 + 4 + 5 + … + 1000 + 1001) - (2 + 4 + … + 1000)

The first group of terms is exactly the sum ∑(1001). Each term in the second grouped sum has a common factor of 2, which we can pull out to get

2 (1 + 2 + … + 500)

so this other group is also a function of ∑(500), and so

S' = ∑(10001) - 2 ∑(500) = 251,001

However, we want to use Gauss' method. We have a sum of the first 501 odd integers. (How do we know there 501? Starting with k = 1, any odd integer can be written as 2k - 1. Solve for k such that 2k - 1 = 1001.)

S' = 1 + 3 + 5 + … + 997 + 999 + 1001

S' = 1001 + 999 + 997 + … + 5 + 3 + 1

2S' = 501 × 1002

S' = 251,001

c/d. I think I've demonstrated enough of Gauss' approach for you to fill in the blanks yourself. To confirm the values you find, you should have

3 + 6 + 9 + … + 300 = 3 (1 + 2 + 3 + … + 100) = 3 ∑(100) = 15,150

and

4 + 8 + 12 + … + 400 = 4 (1 + 2 + 3 + … + 100) = 4 ∑(100) = 20,200

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2 years ago

I'm having trouble on the problem...Will someone please give me step by step?

Reptile [31]