Answer:
<h3>30 tiles need to buy for same cost in both store.</h3>
Step-by-step explanation:
Let x be the number of tiles to buy for same cost in both store.
18+0.69 · n = 1.29 · n,
1.29n -0.69n =18 or
0.6n =18
n =18/0.6=30
Therefore, 30 tiles need to buy for same cost in both store.[Ans]
Now, let's say, we add "x" lbs of the 60% gold alloy, so.. how much gold is in it? well, is just 60%, so (60/100) * x, or 0.6x.
likewise, if we use "y" lbs of the 40% alloy, how much gold is in it? well, 40% of y, or (40/100) * y, or 0.4y.
now, whatever "x" and "y" are, their sum must be 12.4 lbs.
we also know that the gold amount in each added up, must equal that of the 50% resulting alloy.


how much of the 40% alloy? well, y = 12.4 - x.
Total distance traveled = 4 + 6 = 10 miles
Total time = 15 + 25 = 40 minutes
Speed = 10 miles ÷ 40 minutes = 0.25 mph
Answer:

Step-by-step explanation:
we are given surface area and the length of the square base
we want to figure out the Volume
to do so
we need to figure out slant length first
recall the formula of surface area

where B stands for Base area
and P for Base Parimeter
so

now we need our algebraic skills to figure out s
simplify parentheses:

reduce fraction:

simplify multiplication:

cancel 64 from both sides;

divide both sides by 16:

now we'll use Pythagoras theorem to figure out height
according to the theorem

substitute the value of l and s:

simplify parentheses:

simplify squares:

cancel 16 from both sides:

square root both sides:

recall the formula of a square pyramid

where A stands for Base area (l²)
substitute the value of h and l:

simplify multiplication:

reduce fraction:

hence,
