Answer:
1 bag
Step-by-step explanation:
because the other ones are more expensive
Answer:
(d + 3) would be 10
Step-by-step explanation:
<u>Step 1: Distribute</u>



<u>Step 2: Subtract 13.5 from both sides</u>


<u>Step 3: Divide both sides by 4.5</u>


<u>Step 4: Determine what (d+3) is</u>



Answer: (d + 3) would be 10
Answer:
LCM = 20, GCF = 9
Step-by-step explanation:
Suppose Jenny publishes X post per day, and Antonio publishes Y post per day.
So, the combined number of posts will be X+Y multiplied by the number of days.
For example, if the numbers of days is Z, Jenny and Antonio published Z(X+Y) posts.
9514 1404 393
Answer:
x = 4
Step-by-step explanation:
I like to put these in the form f(x) = 0. We can do that by subtracting the right side. Common factors can be cancelled from numerator and denominator, provided they are not zero.

_____
If you leave the numerator as (x-3)(4-x), then there are two values of x that make it zero. Because x=3 makes the equation "undefined", it cannot be considered to be a solution.