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nasty-shy [4]
3 years ago
11

- 63. (26+5) - (86+6) + 3 (6-2)

Mathematics
2 answers:
erastovalidia [21]3 years ago
5 0
I think it’s -2054 but I’m not positive
Radda [10]3 years ago
5 0
- 63. (26+5) - (86+6) + 3 (6-2)

=-2033

The answer is -2033
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Brett writes a number that is the same distance from o as 6 on a number line but in the opposite direction. What number did bret
photoshop1234 [79]
Well since 6 would be in the oppposite direction it would be turned around. Well this leads to non than any other number 9.

Answer ~ 9
4 0
2 years ago
Please answer quick thank you!!
Rudiy27

Answer:

y=\frac{x+9}{n}

Step-by-step explanation:

x/y=n-9

x= y(n-9)

x= yn-9n

-yn= -x-9n

yn= x+9n

y=\frac{x+9}{n}

3 0
2 years ago
Read 2 more answers
Wich one is it???????????
Agata [3.3K]

Answer: Division Property of Equality

4 0
2 years ago
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Help answering this simple Series Question? I don't know what I did wrong!
Tatiana [17]
Looks like you just evaluated the summand for the given value of n, whereas the question is asking you to find the value of the sum for the first n terms.

Let S_k=\displaystyle\sum_{n=1}^k\frac3{(-2)^n}. Then S_k is the kth partial sum.

S_1 happens to be the first term in the series, which is why that box is marked correct:

S_1=\displaystyle\sum_{n=1}^1\frac3{(-2)^n}=\frac3{(-2)^1}=-1.5

But the next partial sum is not correct:

S_2=\displaystyle\sum_{n=1}^2\frac3{(-2)^n}=\frac3{(-2)^1}+\frac3{(-2)^2}=-0.75

and this is not the same notion as the second term (which indeed is 0.75) in the series.
5 0
2 years ago
Let f(x)=3x+5 and g(x)=x^2 find (f-g)(x)
dlinn [17]

Answer:

(f - g)(x) = -x² + 3x + 5

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Equality Properties

<u>Algebra I</u>

  • Function Notation
  • Combining Like Terms

Step-by-step explanation:

<u>Step 1: Define</u>

f(x) = 3x + 5

g(x) = x²

(f - g)(x) is f(x) - g(x)

<u>Step 2: Find (f - g)(x)</u>

  1. Substitute:                   (f - g)(x) = 3x + 5 - x²
  2. Rewrite:                       (f - g)(x) = -x² + 3x + 5
7 0
2 years ago
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