The complete question in the attached figure
Part A) How much sand is currently in the container?
[sand currently in the container]=(5)*(4 1/2)*(2.25)-----> (5)*(4.5)*(2.25)
[sand currently in the container]=50.625 in³
the answer Part a) is 50.625 in³Part B) How much more sand could the container hold before?
[sand could the container hold before]=[5*4.5*3]-[50.625]
[sand could the container hold before]=[67.5]-[50.625]------> 16.875 in³
the answer Part B) is 16.875 in³
Part C) What percent of the container is filled with sand?
the volume of container is [5*4.5*3]=67.5 in³
the volume filled with sand=50.625 in³
therefore
if 100%----------------> 67.5 in³
X---------------------> 50.625 in³
X=(50.625*100)/67.5=75 %
the answer Part C) is 75%
Answer:
76
Step-by-step explanation:
Approximately 6 hours and 18 minutes
Step-by-step explanation:
x = one odd number
the second odd number is (x+2), as both are consecutive.
so,
x × (x + 2) = 195
x² + 2x = 195
x² + 2x - 195 = 0
the formula to solve this is
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = 1
b = 2
c = -195
x = (-2 ± sqrt(4 - 4×1×-195))/(2×1) =
= (-2 ± sqrt(4 + 4×195))/2 = (-2 ± sqrt(784))/2 =
= (-2 ± 28)/2 = -1 ± 14
x1 = -1 + 14 = 13
x2 = -1 - 14 = -15 which is invalid as the numbers are positive.
so, the two numbers are 13 and 15.