(a − 1/a)^2 − (a + 1/a)^2 = (A) 4 (B) -4 (C) 2 (D) -2 (E) 2a
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The tens digit is 3; the units digit is 3-1 = 2; and the number itself is 32.
<em>Lets simplify the problem,</em>
Let assume the "tens" digit be x
Then the "units" difference = (x-1), according to the condition.
Hence, the number itself is N = 10x + (x-1).
Then the number N+8 is 10x + (x-1) + 8 = 10x + x + 7 = 11x + 7.
From the last statement of the problem, we have this equation
Simplify and find "x"
.
Thus the tens digit is 3; the units digit is 3-1 = 2; and the number itself is 32.
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Observe attached picture.
On picture we have:
A = height of flagpole = x ft
B = length of flagpole's shadow = 24 ft
C = height of sign = 6 ft
D = length of sign's shadow = 3 ft
When we draw a picture representing this problem we can also add another line marked in red. This way we can see that we have two right-angle triangles. We can see that both have same angle marked with α.
We can apply trigonometry rules to find height of flagpole.
From small triangle containing sign we can find tangens function:
Similarly we can do for large triangle containing flagpole:
We see that these two equations have same left sides. This means that their right sides must also be same:
We can solve for A:
Height of flagpole is 48 feet.
On picture we have:
A = height of flagpole = x ft
B = length of flagpole's shadow = 24 ft
C = height of sign = 6 ft
D = length of sign's shadow = 3 ft
When we draw a picture representing this problem we can also add another line marked in red. This way we can see that we have two right-angle triangles. We can see that both have same angle marked with α.
We can apply trigonometry rules to find height of flagpole.
From small triangle containing sign we can find tangens function:
Similarly we can do for large triangle containing flagpole:
We see that these two equations have same left sides. This means that their right sides must also be same:
We can solve for A:
Height of flagpole is 48 feet.