Answer:
<2 ~= < 4 - Vertical Angles Theorem
<2 ~= <8 - Alternate Exterior Angles
<4 ~= <8 - Corresponding Angles Theorem
line I || line m - Transitive Property
It would take approximately 30 days for a single-cell amoeba to produce a population of about 10,000 amoebae.
To find out how long it would take for a single-cell amoeba to produce a population of about 10,000 amoebae, we need to calculate the number of times the amoeba doubles. Since the amoeba doubles every 3 days, we can find out the number of doubling periods it would take to reach 10,000 amoebae by dividing 10,000 by 2. This equals approximately 9.965, which means the amoeba would need to double about 9.965 times. Since we can't have a fraction of a doubling period, we can round it up to 10.
Each doubling period is 3 days, so to find out how long it would take, we can multiply the number of doubling periods (10) by the time interval for each doubling period (3 days). 10 x 3 = 30. Therefore, it would take approximately 30 days for a single-cell amoeba to produce a population of about 10,000 amoebae.
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0 = x – 3 – 5x2
0 = 3x – 5 – x2
0 = –3x + 5 – x2
Answer:
Option (a) is correct.
For given values a = 1 , b = -3 , c = -5 the quadratic equation is
Step-by-step explanation:
Given : a = 1 , b = -3 , c = -5
We have to write the quadratic equation having a = 1 , b = -3 , c = -5 and choose the correct option.
The standard form of quadratic equation is , where a, b, c are constant integers.
Given : a = 1 , b = -3 , c = -5
Then Substitute, we get,
Thus, the obtained quadratic equation is same as option (a)
Thus, For given values a = 1 , b = -3 , c = -5 the quadratic equation is
Answer:
a
Step-by-step explanation:
Answer:
The speed is 5t^3 - 10t^2 + 10t - 20 + 46/(t+2) miles per minute
Step-by-step explanation: