H(t) = f(t) - g(t)
= 500(1.2)^t - 380(1.15)^t
Taking out the greatest common factor, which is 20:
= 20[(25)(1.2)^t - (19)(1.15)^t]
This is the third choice.
Note that you cannot subtract the bases of the exponents, for example (1.2^t - 1.15^t) cannot be simplified into something like 0.05^t.
Explanation:
Let M be the midpoint of AB. Then CM is the perpendicular bisector of AB. As such, center O is on CM, and OC is a radius (and CM). The tangent is perpendicular to that radius (and CM), so is parallel to AB, which is also perpendicular to CM.
If you need to go any further, you can show that triangles CMA and CMB are congruent, so (linear) angles CMA and CMB are congruent, hence both 90°.
After clarification we see the respective diameters are


We're asked about the ratio

Now we can go through the choices.
A. If a = 4 and b = 6, then cell a is 100 times greater than cell b.

It would be more proper to say the diameter of cell A is 100 times the diameter of cell B, but yes, this is TRUE.
B. If a = 4 and b = 6, then cell b is 100 times greater than cell a.
FALSE. 10 to the minus 4 is 100 times bigger than 10 to the minus 6, so a is bigger.
C. If a = 4 and b = 6, then cell a is 2 times greater than cell b.
FALSE. We already calculated its 100 times bigger.
D. If a = 4 and b = 6, then cell b is 2 times greater than cell a.
FALSE.
Mike tosses 70%
Ike tosses 67%
Both tosses 50%
<span>Which of the following is closest to the probability that Ike's proportion is ringers is higher than Mike's for those tosses?
</span>
P(m) = 70/100
P(i) = 67/100
P(b) = 50/100
= P(b) * P(i)
= 50/100 * 67/100
= 0.335
The correct answer is letter D) 0.3745.
First of all, a bit of theory: since the area of a square is given by

where s is the length of the square. So, if we invert this function we have
.
Moreover, the diagonal of a square cuts the square in two isosceles right triangles, whose legs are the sides, so the diagonal is the hypothenuse and it can be found by

So, the diagonal is the side length, multiplied by the square root of 2.
With that being said, your function could be something like this:
double diagonalFromArea(double area) {
double side = Math.sqrt(area);
double diagonal = side * Math.sqrt(2);
return diagonal;
}