All categories

3 years ago

How many lakhs are there in 2 millions?

Mathematics

1 answer:

nekit [7.7K]3 years ago

3 0

Answer:

2,000,000 - 2 million

20,00,000 - 20 lakhs

You might be interested in

A person bought a safe which can be unlocked by entering a 2-digit

just olya [345]

39 number is the secret code as the unlock code for the safe.

According to the statement

we have given that a some numbers as a code and we have to find the correct code after applying the given conditions.

So, For this purpose

The given number set is

5 17 29 22 11 26 27 35 39 41

So,

A. One condition : It is an odd number

so, the number set become

5 17 29 11 27 35 39 41

And

B. Second condition : It is greater than 20

so, the number set become

29 27 35 39 41

And

C. Third condition : It is smaller than 40

so, the number set become

29 27 35 39

And

D. Fourth condition : It is not a prime number

so, the number set become

27 35 39

And

E. Fifth condition : It has 13 as one of its factors

so, the number set become

39.

So, 39 number is the secret code as the unlock code for the safe.

Learn more about number set here

brainly.com/question/2689808

#SPJ1

8 0

2 years ago

I need help with part b. I feel like there’s a catch, I want to do the first derivative test, however, I feel like there is a be

Sladkaya [172]

Answer:

The fifth degree Taylor polynomial of g(x) is increasing around x=-1

Step-by-step explanation:

Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

$P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}$

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: $g'(-1)\,(1)$ since the derivative of (x+1) is one,

3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero

Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: $g'(-1)= 7$ as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1