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
Here are the steps
1: Put the compass on Q and make the width equal to the distance from Q to L. Extend line LM towards the left side of L and draw an arc hitting the line segment on the left side of L
2. <span> Without changing the width and position of the compass, draw an arc between L and M.
3. Without changing the width of the compass, put the compass on the point of intersection of the arc and line LM (left side of L). Draw an arc above line LM.
4. Without changing the width of the compass, put the compass on the point of intersection of the arc and line LM (right side of L). Draw an arc above line LM.
5. Use a straight edge to make a line from the intersection of the two arcs above line LM to Q intersecting through L along the way. </span>
Complementary and adjacent
Median is half of add the sum ofthe base and first side
17+5x-14=3+5x
divide it by 2
3+5x ÷2=median
(3+5x)/2=x+12
3+5x=2x+24
3x=21
x=7
Answer is D=7
V=length times width times height
wait, we need it in yards
3ft=1yard
so height=1
v=45 times 30 times 1
v=1350 cubic yards
1 cubic yard=201.974 gallons
so times both sides by 1350
272664.9 gallons
per every 5000 gallons of water we have 1 of chlorine
so 272664.9/5000=54.53298
so we need about 54.5 gallons or 55 gallons of cholrine
