Let the numbers be n, n+2, n+4
Sum equals too= 13+2(n+4), which is 2n+21
a) Equation--> n+n+2+n+4= 2n+21
b) Solution--> 3n+6= 2n+21
=> n= 15
c) Second number--> 17 (15+2)
Third number--> 19 (15+4)
d) 15+15+2+15+4=30+21
=> 51= 51
So, the equation is true.
And pls mark me brainliesttt :)))
Answer:
The answer is 43 and 19 over 21
Step-by-step explanation:
21 goes into 922 43 times. This giving you 903. 43 Will then be your Whole number. You the subtract 922 by 903, this giving you 19. 19 is going to be your numerator. Then you always keep the denomonator which is 21. This giving you 43 and 19 over 21
Answer:
False
Step-by-step explanation:
There are two concepts:
1. Row Echelon Form: There can be more than two <em>row echelon forms</em> of a single matrix, so different sequences of row operations can lead to different <em>row echelon forms</em> of a single matrix.
2. Reduced Row Echelon Form: It's unique for each matrix, so different sequences of row operations always lead to the same <em>reduced row echelon form</em> for the same matrix.
Answer:
Money
Step-by-step explanation:
This is really dumb, but a person can be in debt by 20 cents and another person by 10 cent; therefore, the person who is in debt by 10 cent obviously has to pay less than the person with the 20-cent in debt. Given these facts, the person in debt with one cent has more money then the person in debt with 20 cents.