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.
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The equation between the number of days remaining and number of hours is given as y= -5x+50.
<h3>How to illustrate the equation?</h3>
Let y be the number of days remaining and x be the no.of hours.
From the graph the slope of the line is given by m.
m=(y2 - y1)/(x2 - x1)
m=(30 - 40)/(4 - 2)
m= -10/2 = -5.
The above slope is in between points 2 and 4. As the line has a constant rate of growth the slope is the same between any points of the line.
Therefore the equation between the no.of days remaining and no.of hours is given by
y=mx +c.
y=-5x+c
Now to find the value of c. This can be determined by substituting the x and y values at a particular instant. For x=0,y=50. Therefore by these values, we get,
50= -5(0)+c
c=50.
Therefore the relation between the number of days remaining and number of hours is given as y= -5x+50.
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Answer:
The 4th graph
Step-by-step explanation:
To determine which graph corresponds to the f(x) = \sqrt{x} we will start with inserting some values for x and see what y values we will obtain and then compare it with graphs.
f(1) = \sqrt{1} = 1\\f(2) = \sqrt{2} \approx 1.41\\f(4) = \sqrt{4} = 2\\f(9) = \sqrt{9} = 3
So, we can see that the pairs (1, 1), (2, 1.41), (4, 2), (3, 9) correspond to the fourth graph.
Do not be confused with the third graph - you can see that on the third graph there are also negative y values, which cannot be the case with the f(x) =\sqrt{x}, the range of that function is [0, \infty>, so there are only positive y values for f(x) = \sqrt{x}
Answer:
Time taken for the ball to hit the ground back = 3.08 s
Step-by-step explanation:
h(t)= -16t² + 48t + 4
when will rhe object come back to hit rhe ground?
When the ball is at the level.of the ground, h(t) = 0.
0 = -16t² + 48t + 4
-16t² + 48t + 4 = 0
Solving the quadratic equation
t = 3.08 s or t = -0.08 s
Since the time cannot be negative,
Time taken for the ball to hit the ground back = 3.08 s
Hope this Helps!!!
Step-by-step explanation:
