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Otrada [13]
3 years ago
7

F(x)=2x^2 -5 , simplify f(x+h)/h , where h≠0

Mathematics
1 answer:
7nadin3 [17]3 years ago
4 0

Answer:

\huge\boxed{\dfrac{f(x+h)}{h}=\dfrac{2x^2+4xh+2h^2-5}{h}}

\huge\boxed{\dfrac{f(x+h)}{h}=\dfrac{2x^2-5}{h}+4x+2h}

Step-by-step explanation:

f(x)=2x^2-5\\\\\dfrac{f(x+h)}{h}=\dfrac{2(x+h)^2-5}{h}=\dfrac{2(x^2+2xh+h^2)-5}{h}\\\\\dfrac{f(x+h)}{h}=\dfrac{2x^2+4xh+2h^2-5}{h}\\\\h\neq0

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Tamara bought 4 pack of move tickets $42.50 .How much does the single ticket cost. (3)which digit is in the hundreds
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Than rounding that up to the nearest hundreds which is the number 2, the total is $10.63.
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8 0
3 years ago
Read 2 more answers
I need help ASAP! Can anyone please check my work?
STALIN [3.7K]

A = event the person got the class they wanted

B = event the person is on the honor roll

P(A) = (number who got the class they wanted)/(number total)

P(A) = 379/500

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There's a 75.8% chance someone will get the class they want

Let's see if being on the honor roll changes the probability we just found

So we want to compute P(A | B). If it is equal to P(A), then being on the honor roll does not change P(A).

---------------

A and B = someone got the class they want and they're on the honor roll

P(A and B) = 64/500

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P(B) = 144/500

P(B) = 0.288

P(A | B) = P(A and B)/P(B)

P(A | B) = 0.128/0.288

P(A | B) = 0.44 approximately

This is what you have shown in your steps. This means if we know the person is on the honor roll, then they have a 44% chance of getting the class they want.

Those on the honor roll are at a disadvantage to getting their requested class. Perhaps the thinking is that the honor roll students can handle harder or less popular teachers.

Regardless of motivations, being on the honor roll changes the probability of getting the class you want. So Alex is correct in thinking the honor roll students have a disadvantage. Everything would be fair if P(A | B) = P(A) showing that events A and B are independent. That is not the case here so the events are linked somehow.

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Whats the difference between linear equations and quadratic equation
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Answer:

Step-by-step explanation:

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