I think the answer is.e=mc3
B) ABCD is a rhombus as all the sides are congruent (equal)
Answer: 35
Step-by-step explanation:
Given : A bag contains four red marbles, two green ones, one lavender one, three yellows, and one orange marble.
Total = 4+2+1+3+1=11
To find sets of four marbles include none of the red ones, we need to exclude red marbles when we count the total number of marbles.
Then, the total marbles(exclude red) =11-4=7
Now, the combination of 7 marbles taking 4 at a time is given by :-
![^7C_4=\dfrac{7!}{4!(7-4)!}=\dfrac{7\times6\times5\times4!}{4!3!}=35](https://tex.z-dn.net/?f=%5E7C_4%3D%5Cdfrac%7B7%21%7D%7B4%21%287-4%29%21%7D%3D%5Cdfrac%7B7%5Ctimes6%5Ctimes5%5Ctimes4%21%7D%7B4%213%21%7D%3D35)
Hence, the number of sets of four marbles include none of the red ones = 35
Answer:
The sequence is:
10, 30, 50, 70, 90.....................
Step-by-step explanation:
We have,
First term (a) = 10
Common difference (d) = ?
Sum of first 5 terms (
) = 250
or, ![\frac{n}{2} [{2a+(n-1)d}] = 250](https://tex.z-dn.net/?f=%5Cfrac%7Bn%7D%7B2%7D%20%5B%7B2a%2B%28n-1%29d%7D%5D%20%3D%20250)
or, ![\frac{5}{2} [2*10 + 4d]=250](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B2%7D%20%5B2%2A10%20%2B%204d%5D%3D250)
or, ![\frac{5}{2} * 4[5+d]=250](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B2%7D%20%2A%204%5B5%2Bd%5D%3D250)
or, 10(5 + d) =250
or, 5 + d = 25
∴ d = 20
Now,
2nd term = a + d = 10 + 20 = 30
3rd term = a + 2d = 10 + 2*20 = 10 + 40 = 50
4th term = a + 3d = 10 + 3*20 = 10 + 60 = 70
5th term = a + 4d = 10 + 4*20 = 10 + 80 = 90