Jul 17, 2017 · See a solution process below: First, let's give the two numbers a name: Number 1: we will call: n Number 2: we will call: m From the information in the problem we can write these two equations: Equation 1: n + m = 17 Equation 2: n - m = 29 Step 1 Solve the first equation for n: n + m = 17 n + m - color(red)(m) = 17 - color(red)(m) n + 0 = 17 - m n = 17 - m Step 2 Substitute (17 - m) for n in ...

Feb 27, 2018 · See a solution process below: First, let's call "a number": n Then, "the difference of a number and 17" can be written as: n - 17 "Half" of this can be written as: 1/2 xx (n - 17) Or 1/2 * (n - 17) Or 1/2(n - 17) Or (n - 17)/2

Jun 20, 2016 · a_n=17.486" "milligrams The half life =1590" "years t_0=100" "time=0 t_1=50" "time=1590 t_2=25" "time=2(1590) t_3=12.5" "time=3(1590) a_n=a_0*(1/2)^n 1 " period"=1590" " years n=4000/1590=2.51572327 a_n=100*(1/2)^(2.51572327) a_n=17.486" "milligrams God bless...I hope the explanation is useful.

Jun 26, 2018 · 18% ammonia = NH_3 First find the total molecular mass of the molecule. 1 xx N = 1 xx 14 = 14 2 xx H = 2 xx 1 = 3 14 + 3 = 17 seco= % nd find the percent of the total molecular mass is Hydrogen 3/17 = %/100 Solve for the percent multiple both sides by 100 and divide by 17 300/17= % 17.6 to two significant number this is 18%

Jun 15, 2018 · Find cos u cos^2 u = 1 - sin^2 u = 1 - 9/25 = 16/25 cos u = 4/5 (because u lies in Quadrant 4) cos v = 15/17. Find sin v. sin^2 v = 1 - cos^2 v = 1 - 225/289 = 64/289 sin v = 8/17 (because v lies in Quadrant 1) Replace these above values into identity (1) cos (u - v) = (15/17)(4/5) + (-3/5)(8/17) = 36/85

Jul 27, 2018 · full ground state electron configuration: 1s^2 2s^2 2p^6 3s^2 3p^5 abbreviated: "[Ne]" 3s^2 3p^5 Chlorine has an atomic number of 17, which means it has 17 protons and therefore 17 electrons in its atomic form. We'll need to know how many sublevel is present in each energy level, and in turn, how many electrons each sublevel can accommodate. From the given table, for energy level 1, there's ...

How do you write #12.17# in scientific notation? How do you write #0.007 498 3 times 10^4# in scientific notation? How do you write #1764.9# in scientific notation?

Dec 4, 2016 · By realizing that 10 = 7(1) + 3, 17 = 7(2) + 3, etc. An arithmetic sequence is one where some constant difference between numbers in a sequence gives you a pattern to get to the next unlisted number in the sequence. In this case, the difference is 7. When one works backwards and subtracts 7 from 10, the result is 3. At the 0th term, we assume that n > 0 for series and sequences. So, 3 is the ...

Jul 5, 2017 · n th term is a_n= a_1 +(n-1)d The sequence is of arithmatic progression Here 1 st term is a_1=12 , common difference is d=17-12=22-17=5 , n th term is a_n= a_1 +(n-1)d e.g a_5= 12+(5-1)*5= 12=4*5=32 So n th term is a_n= a_1 +(n-1)d [Ans]

Feb 13, 2017 · the nth term is n^2+1 The sequence above is quadratic, and we use the expression an^2+bn+c where 'a' represents 1/2 the second difference and 'c' is the 0th term. We then substitute numbers into the equation to find the value of 'b'. For example, the second difference in your sequence would be 2 because the 1 first differences are 3, 5, 7, 9 and 11. Thus 'a' = 1 because 2/2 = 1. Then we work ...