I used to hate fractions. But in time, you learn to love them. This is because there's a big difference between fractions and decimals, even though when you divide the actual fraction it comes out to a decimal. Decimals go on and on sometimes, and it would be impossible to write out all those numbers, especially when taking a timed test, for example. Fractions, in this case, would be much more useful (as long as you know how to use them to your advantage). Fractions are basically all those decimal numbers wrapped up into a single, simple division. It makes the outcome of your answer much more accurate than if you estimate every decimal you get throughout a math problem. The more you estimate throughout the problem-solving process, the less accurate your final answer will be. Hence why teachers will usually tell you to estimate when you're putting down the final answer. Fractions are complex at times, so it may be easier to use them in decimal form for certain situations (especially if the decimal form is short and sweet). A world without fractions will result in many, many inaccurate situations involving mathematical knowledge.
Answer:
The answer is 3 11/20
Step-by-step explanation:
Subtract whole numbers, 11 - 8 = 3
Then combine fractions 3/4 - 2/10= 11/20
hope this helps! :D
Since g(6)=6, and both functions are continuous, we have:
![\lim_{x \to 6} [3f(x)+f(x)g(x)] = 45\\\\\lim_{x \to 6} [3f(x)+6f(x)] = 45\\\\lim_{x \to 6} [9f(x)] = 45\\\\9\cdot lim_{x \to 6} f(x) = 45\\\\lim_{x \to 6} f(x)=5](https://tex.z-dn.net/?f=%5Clim_%7Bx%20%5Cto%206%7D%20%5B3f%28x%29%2Bf%28x%29g%28x%29%5D%20%3D%2045%5C%5C%5C%5C%5Clim_%7Bx%20%5Cto%206%7D%20%5B3f%28x%29%2B6f%28x%29%5D%20%3D%2045%5C%5C%5C%5Clim_%7Bx%20%5Cto%206%7D%20%5B9f%28x%29%5D%20%3D%2045%5C%5C%5C%5C9%5Ccdot%20lim_%7Bx%20%5Cto%206%7D%20f%28x%29%20%3D%2045%5C%5C%5C%5Clim_%7Bx%20%5Cto%206%7D%20f%28x%29%3D5)
if a function is continuous at a point c, then
,
that is, in a c ∈ a continuous interval, f(c) and the limit of f as x approaches c are the same.
Thus, since
, f(6) = 5
Answer: 5
Answer:
The area of the park is 119,966.14 cm if needed to round it would be 119,966 cm.
Step-by-step explanation:
first we would use A=length x width
so a = 614 cm x 212 cm = 130,168 cm
now we find the area of the pond using a= pi x r^2
the diameter is 114 cm so we cut that in half to find the radius which is 57 cm
a = pi x 57^2
a= 3.14 x 3249
a= 10201.86 cm
now we subtract the area of the pond from the whole area
130,168 - 10201.86 = 119,966.14 cm
The area of the park is 119,966.14 cm if needed to round it would be 119,966 cm
