Directions: Using only numbers 1-6 (and only once per inequality), fill in the boxes to create a true statement with the smalles
t possible interval:
1 answer:
Answer:
Step-by-step explanation:
√2=1.41
So, the smallest possible interval will be 1.40 and 1.42
√3=1.73
So, the smallest possible interval will be 1.72 and 1.74
√4=2.00
So, the smallest possible interval will be 1.99 and 2.01
√5=2.24
So, the smallest possible interval will be 2.23 and 2.25
√6=2.45
So, the smallest possible interval will be 2.44 and 2.46
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If the radio active isotope has the half life of 24100 years, then the initial quantity is 2.16 grams
The half life in years = 24100
Consider the quantity of the radio active isotope remaining
y =
When t = 1000 the y = 1.2
y = C/2 when t = 1599
Substitute the values in the equation
C/2 =
Cancel the C in both side
1/2 =
Here we have to apply ln to eliminate the e terms
ln (1/2) = 24100k
k = ln(1/2) / 24100
k = -2.87× 10^-5
To find the initial value we have to substitute the value of k and y in the equation
1.2 = Ce^{1000 × -2.87× 10^-5}
C = 1.2 / e^(-0.0287)
C = 2.16 gram
Hence, the initial quantity of the radioactive isotope is 2.16 gram
Learn more about half life here
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Answer:
Length of BH is 2 units.
Step-by-step explanation:
We have, ΔABC with ∠ABC = 90° and an altitude BH.
Also, it is given that the length of sides AH is 4 units and HC is 1 units.
Since, the 'Altitude Rule' states that 'the length of the altitude is the geometric mean of the line segments it bisects'.
So, we get that,
i.e.
i.e.
i.e.
Since, the length of the side cannot be negative.
So, length of BH is 2 units.