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goblinko [34]
3 years ago
8

Given the functions: f(x) = x^2-9 and g(x) = 3 what is (f/g)(x)? A. x-3 B. x-9 C. x+3 D. x+9

Mathematics
1 answer:
uysha [10]3 years ago
3 0

Answer:

f + g)(x) = f (x) + g(x)

= [3x + 2] + [4 – 5x]

= 3x + 2 + 4 – 5x

= 3x – 5x + 2 + 4

= –2x + 6

(f – g)(x) = f (x) – g(x)

= [3x + 2] – [4 – 5x]

= 3x + 2 – 4 + 5x

= 3x + 5x + 2 – 4

= 8x – 2

(f × g)(x) = [f (x)][g(x)]

= (3x + 2)(4 – 5x)

= 12x + 8 – 15x2 – 10x

= –15x2 + 2x + 8

\left(\small{\dfrac{f}{g}}\right)(x) = \small{\dfrac{f(x)}{g(x)}}(

g

f

)(x)=

g(x)

f(x)

= \small{\dfrac{3x+2}{4-5x}}=

4−5x

3x+2

My answer is the neat listing of each of my results, clearly labelled as to which is which.

( f + g ) (x) = –2x + 6

( f – g ) (x) = 8x – 2

( f × g ) (x) = –15x2 + 2x + 8

\mathbf{\color{purple}{ \left(\small{\dfrac{\mathit{f}}{\mathit{g}}}\right)(\mathit{x}) = \small{\dfrac{3\mathit{x} + 2}{4 - 5\mathit{x}}} }}

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Answer:

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Step-by-step explanation:

The relation between time, speed, and distance can be used to solve this problem. It can work well to consider just the distance between the drivers, and the speed at which that is changing.

<h3>Separation distance</h3>

Jason got a head start of 20 miles, so that is the initial separation between the two drivers.

<h3>Closure speed</h3>

Jason is driving 10 mph faster than Britton, so is closing the initial separation gap at that rate.

<h3>Closure time</h3>

The relevant relation is ...

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Then the time it takes to reduce the separation distance to zero is ...

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Britton will catch up to Jason after 2 hours. In that time, Britton will have driven (2 h)(75 mi/h) = 150 miles.

__

<em>Additional comment</em>

The attached graph shows the distance driven as a function of time from when Britton started. The distances will be equal after 2 hours, meaning the drivers are in the same place, 150 miles from their starting spot.

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