Answer:
<h2><em>
B. (b+3c)+(b+3c) </em></h2><h2><em>C. </em><em>
2(b)+2(3c)</em></h2>
Step-by-step explanation:
Given this expression 2(b+3c), its equivalent expression is derived by simply opening up the bracket as shown below;
Open the parenthesis by multiplying the constant outside the bracket with all the variables in parenthesis.
= 2(b+3c)
= 2(b)+ 2(3c)
= 2b +2*3*c
= 2b +6c
It can also be written as sum of b+3c in 2 places i.e (b+3c)+(b+3c) because multiplying the function b+3c by 2 means we are to add the function by itself in two places.
<em>Hence the equivalent expression are (b+3c)+(b+3c) and 2(b)+2(3c) or 2b+6c</em>
do it step by step will it work yes
For 
between 
and 
, we have
is continuous over its domain, so the intermediate value theorem tells us that
is true for 
.
For all , we take into account that is 
-periodic, so the above inequality can be expanded to
where 
is any integer. Equivalently,
To get the corresponding solution set for
simply replace with 
:
Okay so what you would do is 1.5 multiply by 3 find out the answer there and then subtract the 5 cups with the answer.
so if it is 1.5×3 that would equal ____ then you would subract 5 - ___ then there is your answer
when
;
for
; and
for