Jasmine is making 150 bracelets and she needs 26 cm of silver wire for each bracelet. She will buy either the 3.7 metre or the 10.5 metre packs. She wants to pay as little as possible for the silver wire. How much will she have to pay for the silver wire to make 150 bracelets? £

The least possible price is  p  =  £110

Step-by-step explanation:

From the question we are told that

     The number of bracelets to be made is  

      The length of silver require for on bracelet is  

      The option of silver length packs that she buys is  a =  10.5 m  packs

                                                                                           b  =  3.7 m packs

    Generally

                   1 bracelet    0.26 m  

                   150 bracelet  z

=>        

=>        

Now for option a  i.e  10.5 m  per pack

    The number of packs require is

=>    

=>      

given that the number of packs cannot be a fraction but an integer hence she needs to purchase v =  4

and that 4 packs would equal  t  =  4 * 10.5 =  42 meters of  silver

 Now for option d  i.e  3.7 meters per pack  

 The number of packs requires is  

=>        

=>        

given that the number of packs cannot be a fraction but an integer hence she needs to purchase w= 11

and that 11 packs would equal  t  = 11 * 3.7 =  40.7 meters of  silver

So the comparing the option and option b we see that for her to pay as little as possible she needs to go for option b since option be will produce the 150  bracelet with a little excess while option a will produce the 150 bracelet with much excess

    Assuming the price for the 3.7 m  pack is    £10

     And the price for the 10.7 pack is  £30

The least possible amount she would pay is  

                 p  =  £110  

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

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!