What is the explicit rule for the following sequence: 10, 13, 16, 19, …?

a) f(n) = 10 + (n - )
b) f(n) = 10 + (n - 1)(3)
c) f(n) = -10 + (n - 1)(3)
d) f(n) = -10 + (n - )

Option B. f(n) = 10 + (n-1)(3)

Step-by-step explanation:

In order to know the correct answer, you need to put some values of n into the equation given.

Let's begin with a value of n = 1. Now, let's discard some options.

If n = 1, means that the bracket (n-1) always become zero (cause 1-1 = 0), this means that no matter what number you multiply or divide, the result will always be zero.

Option C and D can be easily discarted because when you put n = 1, you get the following:

-10 + (1-1)(3) = -10 + (0)(3) = -10 + 0 = -10.

In option D, would be the same, only that instead of using 3, you'll use -3 but -3 * 0 = 0. The final result is -10, and the sequence begins in 10 positive. That's why Option C and D cannot be the correct option.

We have now option A and B, the difference is on the signus of the 3 (option A is negative and option B is positive). Again, let's use logic, both options, when n = 1, will give the same result, 10. However if we use another number different than 1, ex 2, the result would be different in both options:

n = 2.

a) f(2) = 10 + (2-1)(-3) = 10 + (1)(-3) = 10 - 3 = 7.

As you can see in option A, the next number is decreased, it's not 13.

Therefore, the only and correct option would be the b, and here's the proof:

f(1) = 10 + (1-1)(3) = 10 + 0 = 10

f(2) = 10 + (2-1)(3) = 10 + 3 = 13

f(3) = 10 + (3-1)(3) = 10 + 6 = 16

f(4) = 10 + (4-1)(3) = 10 + 9 = 19

the correct answer is x/3+10≥32

hope this !

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length