Radius is the half of diameter.
That means the radius is 2inch/2=1inch.
Good luck mate!
(5raise to power 5)÷(4)=3125/4)
Answer:
450,000 codes
Step-by-step explanation:
Number of digits in the access code = 6
Conditions:
- First digit cannot be 2
- Last digit must be odd
Since, there are 10 digits in total from 0 to 9 and first digit cannot be 2, there are 9 ways to fill the place of first digit.
Last digit must be odd. As there are 5 odd digits from 0 to 9, there are 5 ways to fill the last digit.
The central 4 digits can be filled by any of the 10 numbers. So, each of them can be filled in 10 ways.
According to the fundamental rule of counting, the total possible codes would be the product of all the possibilities of individual digits.
Therefore,
Number of possible codes = 9 x 10 x 10 x 10 x 10 x 5 = 450,000 codes
Hence, 450,000 different codes are possible for the lock box
Let the width be x.
Length = 2x -1
The area of rectangle = 45 cm²
x*(2x -1) = 45
2x² -x = 45
2x² -x - 45 = 0 This is a quadratic equation
comparing to ax² + bx + c = 0, a = 2, b = -1, c = -45
x = (-b + √(b² -4ac)) / 2a or (-b - √(b² -4ac)) / 2a
x = (- -1 + √((-1)² -4*2*-45)) / 2*2 or (- -1 - √((-1)² -4*2*-45)) / 2*2
x = (1 + √(1 +360)) / 2*2 or (1 - √(1 + 360) / 2*2
x = (1 + √361) / 4 or (1 - √361) / 4
x = (1 + 19) / 4 or (1 - 19) / 4
x = 20/4 or -18/4
x = 5 or -4.5
x can't be negative since we are solving for side.
x = 5 as the only valid solution.
Recall, the width = x = 5.
Length = 2x - 1 = 2*5 - 1 = 10 - 1 = 9
Hence the length = 9cm , and width = 5cm
