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yKpoI14uk [10]
2 years ago
9

Solve:2 In 3 = In(x-4)

Mathematics
1 answer:
vovikov84 [41]2 years ago
8 0

x=13 and don't forget to mark me brainlist!

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What are the values of a, b and c in the quadratic equation -2x^2+4x-3=0?
AlladinOne [14]

Answer:

a=-2\\b=4\\c=-3

Step-by-step explanation:

In a quadratic equation in the Standard form

ax^2+bx+c=0

You need to remember that "a", "b" and "c" are the numerical coefficients (Where "a" is the leading coefficient and it cannot be zero: a\neq0).

You can observe that the given quadratic equation is written in the Standard form mentioned before. This is:

-2x^2+4x-3=0

Therefore, you can identify that the values of "a", "b" and "c" are the following:

a=-2\\b=4\\c=-3

4 0
3 years ago
Read 2 more answers
I’ll give brainliest
Nat2105 [25]

Answer:

x = -22

Step-by-step explanation:

6(x+15) = -42

x+15 = -7

x = -22

6 0
3 years ago
I need help QUICK!!!
Flauer [41]

Answer:

87 in³

Step-by-step explanation:

that is the solution above

8 0
2 years ago
At Northside ,40% of the sixth grade students said that hip-hop is their favorite kind of music. If 100 students are at the scho
Anon25 [30]

Answer:

Step-by-step explanation:

Here is the options

6 0
3 years ago
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A manufacturing company has two retail outlets. It is known that 30% of all potential customers buy
aleksandrvk [35]

Answer:

(a) A y P(A) = 0.4 (b) \bar{B} y P(\bar{B})=0.5 (c) \bar{A}∪\bar{B} y P(\bar{A}∪\bar{B}) = 0.9 (d) \bar{A}∩\bar{B} y P(\bar{A}∩\bar{B})=0.2

Step-by-step explanation:

A was defined as the event that a potential customer, randomly chosen, buys from outlet 1 in the original problem statement. We know that B denotes the event that a randomly chosen customer buys from outlet 2. So

P(A\cap \bar{B}) = 0.3, P(B\cap \bar{A}) = 0.4 and P(A\cap B) = 0.1

(a) P(A) = P(A\cap (B\cup\bar{B})) = P(A\cap B) + P(A\cap \bar{B}) = 0.1 + 0.3 = 0.4

(b)  P(B) = P(B\cap (A\cup\bar{A})) = P(B\cap A) + P(B\cap \bar{A}) = 0.1 + 0.4 = 0.5

P( \bar{B}) = 1-P(B) = 1-0.5 = 0.5

(c) The customer does not buy from outlet 1 is the complement of A, i.e.,  \bar{A}, and the customer does not buy from outlet 2 is the complement of B, i.e.,  \bar{B}, so, the customer does not buy from outlet 1 or does not buy from outlet 2 is  \bar{A}∪ \bar{B} and P(\bar{A}∪ \bar{B}) = P((A\cap B)^{c}) by De Morgan's laws

P((A\cap B)^{c})  = 1-P(A∩B)=1-0.1=0.9

(d) The customer does not buy from outlet 1 is the complement of A, and the customer does not buy from outlet 2 is the complement of B, so we have that the statement in (d) is equivalent to \bar{A}∩\bar{B} and P( \bar{A}∩\bar{B}) = P((AUB)^{c}) by De Morgan's laws, and

P((AUB)^{c}) = 1-P(A∪B)=1-[P(A)+P(B)-P(A∩B)]=1-[0.4+0.5-0.1]=1-0.8=0.2

8 0
3 years ago
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