Round off to 4 decimal places

in Other Math Topics by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

Plug in the given values:


Let f(h)=3[9cos⁻¹((3-h)/3)-(3-h)√(6h-h²)]-14.

We need to identify two values, a and b, of h such that f(a)<0 and f(b)>0.

We can limit a and b by noting that 6h-h²≥0, that is, h(6-h)≥0, because the square root only applies to positive quantities. Therefore 0≤h≤6. The inverse cosine function also has a restricted argument which gives us the same range for h. Let a=0, b=6 and f(0)=-14 and f(6)=70.823 approx.

Applying the Regula-Falsi formula for the root:

h=af(b)-bf(a))/(f(b)-f(a))=-6f(0)/(f(6)-f(0))=84/(70.823+14)=0.9903 approx.

f(0.9903)<0, so we know that the root lies between 0.9903 and 6. Readjust a to 0.9903 and leave b=6.

Next estimate for h:



f(1.3106)<0, so the root now lies between 1.3106 and 6. Readjust a to 1.3106 and leave b=6.

We continue this procedure until we reach a value of h which is stable to a sufficient number of decimal places and we end up with h=1.33096242.

Rounded to 4 decimal places this is h=1.3310.


by Top Rated User (840k points)

Related questions

Welcome to, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
86,310 questions
92,368 answers
23,929 users