The width used for the car spaces are taken as a multiples of the width of
the compact car spaces.
Correct response:
- The store owners are incorrect
<h3 /><h3>Methods used to obtain the above response</h3>
Let <em>x</em><em> </em>represent the width of the cars parked compact, and let a·x represent the width of cars parked in full size spaces.
We have;
Initial space occupied = 10·x + 12·(a·x) = x·(10 + 12·a)
New space design = 16·x + 9×(a·x) = x·(16 + 9·a)
When the dimensions of the initial and new arrangement are equal, we have;
10 + 12·a = 16 + 9·a
12·a - 9·a = 16 - 10 = 6
3·a = 6
a = 6 ÷ 3 = 2
a = 2
Whereby the factor <em>a</em> < 2, such that the width of the full size space is less than twice the width of the compact spaces, by testing, we have;
10 + 12·a < 16 + 9·a
Which gives;
x·(10 + 12·a) < x·(16 + 9·a)
Therefore;
The initial total car park space is less than the space required for 16
compact spaces and 9 full size spaces, therefore; the store owners are
incorrect.
Learn more about writing expressions here:
brainly.com/question/551090
Hello there!
So, if this turtle is only going 2.5 miles per hour. This would mean that (EACH) hour, this turtle only arrived 2.5 miles. So, we do,
We would have to count how many ("2.5's" they are above.)
We count only
4.
I would take this turtle
4 miles to swim 10 kilometers.
So in this problem we need to do some addition.
take the starting balance (-200) and add it to the amount you paid back (30)
-200 + 30 = -170
So first I would say, what if all of them were dimes, how far away would it be from $14?
So 92 coins * 10 cents = $9.20
So it's 4.80 dollars away from 14 dollars.
So if we were to switch one to a quarter, it would increase by 0.15 cents.
So we want to see how many increases we need to reach 4.80 dollars more.
4.80/0.15 = 32
So there are 32 quarters and 60 dimes.
