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
9514 1404 393
Answer:
Step-by-step explanation:
Arc DV is 100°, so any inscribed angle that intercepts that arc will have a measure that is half of 100°, or 50°.
Inscribed angles DCV and VBD both intercept arc DV, so both are 50°.
Solve for p by simplifying both sides of the equation, then isolation the variable
p = 1 + m/2
Hope this helps! :)
Answer:
50 students from each grade
Step-by-step explanation:
If we get the similar amount of people from each and every grade so this would represent the percentage occur from each grade this determine the number of people who vote per grade and represent that how much would the vote for student comes
Also 50 is easily divisible by 750
Therefore the above represent the answer
