Answer:
The half life of the car is 3.98 years.
Step-by-step explanation:
The value of the car after t years is given by the following equation:
In which V(0) is the initial value and r is the constant decay rate, as a decimal.
The value of a certain car decreases by 16% each year.
This means that
So
What is the 1⁄2-life of the car?
This is t for which V(t) = 0.5V(0). So
The half life of the car is 3.98 years.
The perimeter of a triangle is the sum of the lengths of its sides. The length of each side can be found using the Pythagorean theorem. Effectively, each pair of points is treated as the end-points of the hypotenuse of a right triangle with legs parallel to the x- and y-axes. The leg lengths are the differences betweeen the x- and y- coordinates of the points.
The difference of the x-coordinates of segment AB are 0-(-3) = 3. The y-coordinate difference is 3-(-1) = 4. So, the leg lengths of the right triangle whose hypotenuse is segment AB are 3 and 4. The Pythagorean theorem tells us
... AB² = 3² +4² = 9 +16 = 25
... AB = √25 = 5
You may recognize this as the 3-4-5 triangle often introduced as one of the first ones you play with when you learn the Pythagorean theorem.
LIkewise, segment AC has coordinate differences of ...
... C - A = (1, 2) -(-3, -1) = (4, 3)
These are the same leg lengths (in the other order) as for segment AB, so its length is also 5.
Segment BC has coordinate differences ...
... C - B = (1, 2) -(0, 3) = (1, -1)
The length of the line segment is figured as the root of the sum of squares, even though one of the coordinate differences is negative. The leg lengths of the right triangle used for finding the length of BC are the absolute value of these differences, or 1 and 1. Then the length BC is
... BC = √(1² +1²) = √2 ≈ 1.4
So the perimeter of the triangle ABC is
... AB + BC + AC = 5 + 1.4 + 5 = 11.4 . . . . perimeter of ∆ABC in units
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Please be aware that the advice to "round each step" is <em>bad advice,</em> in general. For real-world math problems, you only round the final result. You always carry at least enough precision in the numbers to ensure that there will not be any error in the final rounding.
In this problem, the only number that is not an integer is √2, so it doesn't really matter.
