<u>Answer-</u>
At
the curve has maximum curvature.
<u>Solution-</u>
The formula for curvature =

Here,

Then,

Putting the values,

Now, in order to get the max curvature value, we have to calculate the first derivative of this function and then to get where its value is max, we have to equate it to 0.

Now, equating this to 0






Solving this eq,
we get 
∴ At
the curvature is maximum.
B. THE 2 IN THE TENTHS PLACE IS 10 TIMES THE VALUE OF THE 2 IN THE HUNDREDTHS PLACE.
2 2 . 2 2 2
tens ones decimal point tenths hundredths thousandths
2 in the tenths place = 0.20
2 in the hundredths place = 0.02
0.02 x 10 = 0.20
Answer:
A
Step-by-step explanation:
Because I know how to do it
8 / 6 3/10
= 8 * 10/63
=80/63
3 4/15 / 80/63
= 49/15 * 63/80
=<u>2.5725</u>
So, We Have A Rate That We Need To Simplify. We Have:
88 students for every 4 classes
So, We Need To Simplify This Rate. In Order To Do This, We Need To Change Is To A Fraction. It Is:
88 students
---------------
4 classes
Now, We Have To Simplify. We Can Do That By Remembering How To Simplify Fractions.
So,
88 ÷ 2 = 44 ÷ 2 = 22 students
--- --- ---------------
4 ÷ 2 = 2 ÷ 2 = 1
So, The Unit Rate For 88 Students For 4 Classes Is:
22 Students For One Class