A pendulum hangs from a clock and swings over its base.

The distance between the pendulum and the base of the

clock, f(x), over the horizontal distance, x, is modeled by

the function shown.

f(x) = 0.25x² – 2x + 10

What is the minimum distance between the pendulum and

the base of the clock?

XI XII

4.2

Step-by-step explanation:

Given the distance between the pendulum and the base of the

clock, f(x), over the horizontal distance, x, modeled by

the function shown.

f(x) = 0.25x² – 2x + 10

In order to get the minimum distance between the pendulum and the base of the clock, first, we need to find the turning point of the function.

At turning point, d{f(x)}/dx = 0

d{f(x)}/dx = 0.5x-2 = 0

0.5x = 2

x = 2/0.5

x = 4

Then we will substitute x = 4 into the function as shown;

f(4) = 0.25(4)²-2(0.4)+10

f(4) = 4-9.8+10

f(4) = 4.2

The minimum distance between the pendulum and the base of the clock is 4.2

13

step-by-step explanation:

169-1=168

168-3=165

165-5=160

160-7=153

153-9=144

144-11=133

133-13=120

120-15=105

105-17=88

88-19=69

69-21=48

48-23=25

25-25=0

since it took 13times to completely subtract it hence its square root is 13