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vampirchik [111]
3 years ago
12

F(x)=24/3x-2 find f(-2)

Mathematics
1 answer:
Anna007 [38]3 years ago
4 0
F(x)=24/3x-2 ==> substitute -2 for x
f(-2)=24/3(-2)-2===> Multiply -2 by 24/3
f(-2)=-48/3-2==> Divide -48 by 3
f(-2)=-16-2 ===>Combine
f(-2)=-18
Therefore f(-2) is -18
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Carlota wrote the equation y + 1 = 2 (x - 3) for the line passing through the points (-1, 3) and (2, 9). Explain and correct her
Art [367]

For this case we have the following equation:


y + 1 = 2 (x - 3)

That can be rewritten in the form y = mx + b

Where:


  • m is the slope of the line
  • b is the cut point

So, we have:


y + 1 = 2 (x - 3)\\y + 1 = 2x-6\\y = 2x-6-1\\y = 2x-7

Where:


m = 2 is the slope


b = -7 is the cut point


Carlota has the following points:


(-1, 3) and (2, 9)


To know if the line y = 2x-7 passes through these points, we must replace them in the equation and the equality must be fulfilled. So:


Point (-1, 3):


Substituting:


3 = 2 (-1) -7\\3 = -2-7

3 = -9 It's false, equality is not met. The point (-1, 3) does not go through the line.


The equation written by Carlota is erroneous, the procedure to follow is:


Given(x1, y1) = (- 1, 3) and(x2, y2) = (2, 9), we find the slope:


m=\frac{(y2-y1)}{(x2-x1)}

m=\frac{(9-3)}{(2-(-1))}

m=\frac{6}{3}

m=2

We observe that the slope found by Carlota is the same. Let's see cut point "b". For this we substitute any of the points given in the equation:


y = 2x + b

Substituting (2,9) we have:


9 = 2 (2) + b\\9 = 4 + b\\b = 9-4\\b = 5

Thus, Carlota's error was at the cut-off point. The correct equation of the line that passes through the given points is y = 2x + 5

Answer:


The correct equation of the line that passes through the given points is y = 2x + 5

Carlota's mistake was at the cutoff point


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3 years ago
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klemol [59]
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Because I just multiplied 12 by 5/2 which is 30
Then I multiplied 3 by 5/2 which is 7.5
6 0
4 years ago
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Rina8888 [55]

Given:

The graph of a function y=h(x).

To find:

The interval where h(x)>0.

Solution:

From the given graph graph it is clear that, the function before x=0 and after x=3.6 lies above the x-axis. So, h(x)>0 for x and x>3.6.

The function between x=0 and x=3.6 lies below the x-axis. So, h(x) for 0.

Now,

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For 0, the graph of h(x) is below the x-axis. So, h(x).

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3 0
3 years ago
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Jobisdone [24]

Answer:

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