Well.
The relation between the circumference of a circle and its diameter is described by the following equation:

where d is the diameter of the circle.
For a giving diameter (d=3ft), we can substitute in the previous equation
The circumference of a circule= 3.14*3 (you can use ur calculator to find the exact number i guess it will be about 9.42 ft)
I hope it helps.
Answer:
x = -7
Step-by-step explanation:
Recall this rule of exponents:
a^b*a^c = a^(b+c)
We have:
6^9*6^x = 6^2
Then 9 + x = 2. Subtracting 9 from both sides, we get: x = -7
Answer:
Step-by-step explanation:
f(x) = x⁵ – 8x⁴ + 16x³
As x approaches +∞, the highest term, x⁵, approaches +∞.
As x approaches -∞, x⁵ approaches -∞ (a negative number raised to an odd exponent is also negative).
Now let's factor:
f(x) = x³ (x² – 8x + 16)
f(x) = x³ (x – 4)²
f(x) has roots at x=0 and x=4. x=4 is a repeated root (because it's squared), so the graph touches the x-axis but does not cross at x=4.
The graph crosses the x-axis at x=0.
Step-by-step explanation:
