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REY [17]
3 years ago
7

Bicycle license plates in Flatville each contain three letters. The first is chosen from the set ({C,H,L,P,R}) the second from (

A,I,O), and the third from ({D,M,N,T}). When Flatville needed more license plates, they added two new letters. The new letters may both be added to one set or one letter may be added to one set and one to another set. What is the largest possible number of ADDITIONAL license plates than can be made by adding two letters?
Mathematics
1 answer:
amm18123 years ago
7 0
5\cdot3\cdot4=60 - the current possible number of plates

after adding 2 letters into the 1st set:
7\cdot3\cdot4=82
after adding 2 letters into the 2nd set:
5\cdot5\cdot4=100
after adding 2 letters into the 3rd set:
5\cdot3\cdot6=90
after adding 1 letter into the 1st set and 1 into the 2nd set:
6\cdot4\cdot4=96
after adding 1 letter into the 1st set and 1 into the 3rd set:
6\cdot3\cdot5=90
after adding 1 letter into the 2nd set and 1 into the 3rd set:
5\cdot4\cdot4=80

<span>The largest possible number of plates after adding two numbers is 100.
So, </span><span>the largest possible number of additional plates is 100-60=40</span>
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Given a rectangle with length of (2x+9)cm and width of (3x+1)cm.Two squares, each with sides x cm is removed from the rectangle.
il63 [147K]

Answer: The length is 13cm and the width is 7cm

Step-by-step explanation:

For a rectangle of length L and width W, the area is:

A = W*L

In this case we have:

L = (2*x + 9) cm

W=(3*x + 1) cm

Then the area of the rectangle is:

A = (2*x + 9)*(3*x + 1) cm^2

A = (6*x^2 + 2*x + 27*x + 9) cm^2

A = (6*x^2 + 29*x + 9) cm^2

now we remove two squares with sides of x cm

The area of each one of these squares is (x cm)*(x cm)  = x^2 cm^2

Then the area of the figure will be:

area = (6*x^2 + 29*x + 9) cm^2 - (2*x^2 ) cm^2

area = (4*x^2 + 29*x + 9) cm^2

Now we know that the area of this shape is 83 cm^2, then we need to solve:

83 cm^2 = (4*x^2 + 29*x + 9) cm^2

0 =  (4*x^2 + 29*x + 9) cm^2 - 83 cm^2

0 = (4*x^2 + 29*x - 74) cm^2

Then we need to solve:

0 = 4*x^2 + 29*x - 74

Here we can use Bhaskara's equation, the solutions of this equation are given by:

x = \frac{-29 \pm \sqrt{29^2 - 4*4*(-74)}  }{2*4} = \frac{-29 \pm 45}{8}

Then the two solutions are:

x = (-29 - 45)/8 = -9.25  (for how the length and width are defined, we can not have x as a negative number, then this solution can be discarded).

The other solution is:

x = (-29 + 45)/8 = 2

x = 2

Then the length and width of the rectangle are:

Length = (2*2 + 9)cm = 13 cm

Width = (3*2 + 1)cm = 7cm

4 0
2 years ago
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