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nika2105 [10]
3 years ago
10

If f(x) = x-6 and g(x)= 1/2x (x+3), find g(x) * f(x)

Mathematics
1 answer:
sertanlavr [38]3 years ago
5 0

Answer:

Final answer is g\left(x\right)\cdot f\left(x\right)=\frac{\left(x-6\right)}{2x\left(x+3\right)}.

Step-by-step explanation:

given functions are f(x)=x-6 and g\left(x\right)=\frac{1}{2x\left(x+3\right)}.

Now we need to find about what is the value of g\left(x\right)*f\left(x\right).

g\left(x\right)*f\left(x\right) simply means we need to multiply the value of  f(x)=x-6 and g\left(x\right)=\frac{1}{2x\left(x+3\right)}.

g\left(x\right)\cdot f\left(x\right)=\frac{1}{2x\left(x+3\right)}\cdot\left(x-6\right)

g\left(x\right)\cdot f\left(x\right)=\frac{\left(x-6\right)}{2x\left(x+3\right)}

Hence final answer is g\left(x\right)\cdot f\left(x\right)=\frac{\left(x-6\right)}{2x\left(x+3\right)}.

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Make x the subject of the formula <br><br> x/3z = y-2
wlad13 [49]

Answer:

x = 3z(y - 2)

Step-by-step explanation:

Given

\frac{x}{3z} = y - 2 ( multiply both sides by 3z to clear the fraction )

x = 3z(y - 2)

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3 years ago
The number of international tourist arrivals in Russia in 2012 was 13.5% greater than in 2011. The number of international touri
Bess [88]

The value of k to the nearest integer is 21

  • Let the number of international tourist arrivals in Russia in 2011 be k

If the number of international tourist arrivals in Russia in 2012 was 13.5% greater than in 2011 will be expressed as:

k = 24.7 - (13% of 24.7)

k = 24.7 - (0.135 * 24.7)

k = 24.7 - 3.3345

k = 21.3655

Hence the value of k to the nearest integer is 21

Learn more on decrement here: brainly.com/question/24304697

8 0
2 years ago
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3 years ago
Evaluate the factorial expression.<br> 28!<br> 24
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3 years ago
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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