The area of a circle is given by
We can solve this equation for r: divide both sides by pi:
Consider the (positive) square root of both sides:
So, in your case, we have
Answer: B. Benford's law
Step-by-step explanation:
Benford's law created by Simon Newcomb is used to determine the number of times or percentage that a digit will occur in a series or collection of numbers
Answer: Tom's solution is incorrect.
Step-by-step explanation:
When you add 3/6 + 1/2 it equals ?.
3 + 1 = ?
6 + 2 = ?
We must have common denominators to do this problem.
So we must have 6 as our common denominator.
So 2 x 3 = 6. But we have to do the same to the Numerator.
So 1 x 3 = 3.
Now our problem is 3/6 + 3/6 = ?
3 + 3 = 6/6. Or 1
1 does not equal 4/12 or 1/3. So it makes it Incorrect.
What you should say:
Tom is incorrect, it is because 3/6 + 1/2 = 1, 1 does not equal 4/12 or 1/3. So, it makes it reasonable for the correct answer to be Incorrect
Answer/Step-by-step explanation:
Given:
m<EFH = (5x + 1)°
m<HFG = 62°
m<EFG = (18x + 11)°
Required:
1. Value of x
2. m<EFH
3. m<EFG
SOLUTION:
1. Value of x
m<EFH + m<HFG = m<EFG (angle addition postulate)
(5x + 1) + 62 = (18x + 11)
Solve for x using this equation
5x + 1 + 62 = 18x + 11
5x + 63 = 18x + 11
Subtract 18x from both sides
5x + 63 - 18x = 18x + 11 - 18x
-13x + 63 = 11
Subtract 63 from both sides
-13x + 63 - 63 = 11 - 63
-13x = -52
Divide both sides by -13
-13x/-13 = -52/-13
x = 4
2. m<EFH = 5x + 1
Plug in the value of x
m<EFH = 5(4) + 1 = 20 + 1 = 21°
3. m<EFG = 18x + 11
m<EFG = 18(4) + 11 = 72 + 11 = 83°
Answer:
The probably genotype of individual #4 if 'Aa' and individual #6 is 'aa'.
Step-by-step explanation:
In a non sex-linked, dominant trait where both parents carry and show the trait and produce children that both have and don't have the trait, they would each have a genotype of 'Aa' which would produce a likelihood of 75% of children that carry the dominant traint and 25% that don't. Since the child of #1 and #2, #5, does not exhibit the trait, nor does the significant other (#6), then they both must have the 'aa' genotype. However, since #4 displays the dominant trait received from the parents, it is more likely they would have the 'Aa' genotype as by the punnet square of 'Aa' x 'Aa', 50% of their children would have the 'Aa' phenotype.
