f(1) = 4 is given
f(n) = f(n-1)-4 is the recursive step or rule
f(2) = f(2-1)-4 ... replace n with 2
f(2) = f(1) - 4
f(2) = 4-4 ... replace f(1) with 4
f(2) = 0
----------
Repeat for n = 3
f(n) = f(n-1)-4
f(3) = f(3-1)-4
f(3) = f(2) - 4
f(3) = 0 - 4 .... replace f(2) with 0, which is from f(2) = 0 found earlier
f(3) = -4
---------
Repeat for n = 4
f(n) = f(n-1)-4
f(4) = f(4-1)-4
f(4) = f(3) - 4
f(4) = -4-4 ... plug in f(3) = -4
f(4) = -8
<h3>Answer is -8</h3>
R = 12 inches, the radius of the wheel.
The circumference of the wheel is
2πr = 2π(12) = 24π inches.
WHen the wheel makes 15 revolutions, the car will travel 15 times the circumference, which is
15*(24π) = 360π inches = 1131 inches.
Answer: 1131 inches
The equation of a circle is:

Where (xc,yc) are the coordinates of the center.
Center (5, -5)
Radius = 3
(x-5)²+(y+5)²=3²
x²-10x+25+y²+10y+25=9
x²+y²-10x+10y+41=0
<span>There is not a property of "1" for addition, but the identity property in this case uses "0" instead. If a number is increased by 1, the value will necessary be increased or decreased by the amount added or subtracted. However, if the number either adds or subtracts zero, then value will be left unchanged. Examples: 3+1 = 4, but 3+0 = 3,</span>
Answer:

Step-by-step explanation:
we are given half-life of PO-210 and the initial mass
we want to figure out the remaining mass <u>after</u><u> </u><u>4</u><u>2</u><u>0</u><u> </u><u>days</u><u> </u>
in order to solve so we can consider the half-life formula given by

where:
- f(t) is the remaining quantity of a substance after time t has elapsed.
- a is the initial quantity of this substance.
- T is the half-life
since it halves every 140 days our T is 140 and t is 420. as the initial mass of the sample is 5 our a is 5
thus substitute:

reduce fraction:

By using calculator we acquire:

hence, the remaining sample after 420 days is 0.625 kg