Point b lies on the segment ac . find the length of the segment ac, if ab=7.8 cm, bc=25mm. how many different solutions can this problem have? what if we said that point b lies on the line defined by points a and c?

The segment AC can have ONE solution in both cases. The solution can either be 103mmn or 10.3cm for (a) and 5.3cm in (b)

If point B lies on the segment AC, then:

AB + BC = AC

Given the following segments

AB = 7.8cm  = 78mm

BC = 25mm

AC = 78mm + 25mm

AC = 103mm

Converting to cm

10mm = 1cm

103mm = AC

10AC = 103

AC = 103/10

AC = 10.3cm

This shows that AC can have just ONE solution but different units

b) If B is not on the line BC, tyhe  the measure of AC will be expressed as:

AC = AB - BC

AC = 7.8cm - 2.5cm (Note that 25mm = 2.5cm)

AC = 5.3cm

This shows that AC can have just ONE solution but different units in both cases

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a) If B is between A and C, there can only be one solution:

... AC = AB + BC = 7.8 cm + 2.5 cm

... AC = 10.3 cm

b) If B is not on the segment AC, there can only be one solution: C must lie on AB. Then ...

... AC = 7.8 cm - 2.5 cm

... AC = 5.3 cm

This, together with the solution of Part (a) means there are two solutions.

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!