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inna [77]
3 years ago
13

Simplify (4r - s) - (r + 2s) + (3r - s) A.) 6r - 3s B.) 8r C.) 3r - 2s D.) 6r - 4s

Mathematics
1 answer:
docker41 [41]3 years ago
6 0

Answer: D

Step-by-step explanation:

4r-r+3r can be simplified to just 6r

-s-2s-s can be simplified to just -4s

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First pirson to type 1 i will see if i can give brainlist but idk how bot i will try
amm1812

Answer:

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Step-by-step explanation:

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3 0
3 years ago
Hey I need help with this question:<br><br> 6(x-3)^2-26 = 10
ra1l [238]

Answer:

x = 3 ±sqrt(6)

Step-by-step explanation:

6(x-3)^2-26 = 10

Add 26 to each side

6(x-3)^2-26+26 = 10+26

6(x-3)^2 = 36

Divide by 6

6/6(x-3)^2 = 36/6

(x-3)^2 = 6

Take the square root of each side

sqrt((x-3)^2) = ±sqrt(6)

x-3 =  ±sqrt(6)

Add 3 to each side

x-3+3 =3 ±sqrt(6)

x = 3 ±sqrt(6)

7 0
3 years ago
Please help me! :) I will give 15 thank yous! Please help<br> 
Korvikt [17]
80 because if you multiply all them you would get 64,000 then divide it
3 0
4 years ago
Given P(A and B) 0.20, P(A) 0.49, and P(B) = 0.41 are events A and B independent or dependent? 1) Dependent 2) Independent
blsea [12.9K]

Answer:  The correct option is (1) Dependent.

Step-by-step explanation:  For two events, we are given the following values of the probabilities :

P(A ∩ B) = 0.20,   P(A) = 0.49   and    P(B) = 0.41.

We are to check whether the events A and B are independent or dependent.

We know that

the two events C and D are said to be independent if the probabilities of their intersection is equal to the product of their probabilities.

That is,  P(C ∩ D) = P(C) × P(D).

For the given two events A and B, we have

P(A)\times P(B)=0.49\times0.41=0.2009\neq P(A\cap B)=0.20\\\\\Rightarrow P(A\cap B)\neq P(A)\times P(B).

Therefore, the probabilities of the intersection of two events A and B is NOT equal to the product of the probabilities of the two events.

Thus, the events A and B are NOT independent. They are dependent events.

Option (1) is CORRECT.

8 0
4 years ago
there were a total of 125 cars at a car dealership. A salesperson sold 68 of the cars in one month. Write and solve an inequalit
Angelina_Jolie [31]
Let c represent the number of cars that the salesman has left to sell.

      68 + c ≤ 125
68-68 + c ≤ 125 -68
               c≤57
The solution is c ≤ 57. The salesman has at most 57 cars left to sell

Answer:
68 + c ≤ 125; c ≤ 57 cars or less left to sell.
4 0
4 years ago
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