Respuesta :
[tex] \underline{ \huge \mathcal{ Ànswér} } \huge: - [/tex]
Average of b and c is 8, that is
[tex]➢ \: \: \dfrac{b + c}{2} = 8[/tex]
[tex]➢ \: \: b + c = 16[/tex]
[tex]➢ \: \: c = 16 - b[/tex]
now let's solve for average of c and d :
[tex]➢ \: \: \dfrac{c + d}{2} [/tex]
[tex]➢ \: \: \dfrac{16 - b + 3b - 4}{2} [/tex]
[tex]➢ \: \: \dfrac{12 + 2b}{2} [/tex]
[tex]➢ \: \: \dfrac{2(6 + b)}{2} [/tex]
[tex]➢ \: \: b + 6[/tex]
Therefore, the average of c and d, in terms of b is : -
[tex] \large \boxed{ \boxed{b + 6}}[/tex]
[tex]\mathrm{✌TeeNForeveR✌}[/tex]
Answer:
b+6
Problem:
If the average of b and c is 8, and d=3b-4, what is the average of c and d in terms of b?
Step-by-step explanation:
We are given (b+c)/2=8 and d=3b-4.
We are asked to find (c+d)/2 in terms of variable, b.
We need to first solve (b+c)/2=8 for c.
Multiply both sides by 2: b+c=16.
Subtract b on both sides: c=16-b
Now let's plug in c=16-b and d=3b-4 into (c+d)/2:
([16-b]+[3b-4])/2
Combine like terms:
(12+2b)/2
Divide top and bottom by 2:
(6+1b)/1
Multiplicative identity property applied:
(6+b)/1
Anything divided by 1 is that anything:
(6+b)
6+b
b+6
