Hi Mathies, Help with this vectors excercise pls. Givan that OA (segment) = 13x+7y , OB = 5x+12y and CO = -15+12y write down each of the following vectors in its simplest form a) BA = 8x+ 5y (l got it, i ve done it) b) AC= ?? i cant find vector AC thanks in advance

Step-by-step explanation:

Given vector the following vector equations OA = 13x+7y , OB = 5x+12y and CO = -15x+12y, the following expression is true about vector OA, OB and OC;

OA+OB = CO (CO is the resultant since its is moving in the opposite direction compare to OA and OB)

Also BO+OA = BA and AO+OC = AC

If OB = 5x+12y, then BO = -(5x+12y)

BO = -5x-12y (BO = -OB)

Since BO+OA = BA

BA = -5x-12y + 13x+7y

BA = -5x+13x-12y+7y

BA = 8x-5y

Similarly AO+OC = AC

Since AO = -OA and OC = -CO

-OA-CO  = AC

AC = -(13x+7y)-(-15x+12y)

AC = -13x-7y+15x-12y

AC = -13x+15x-7y-12y

AC = 2x-19y

step by step resolutions

1) n+n+1+n+2=51

3n+3=51

3n=48

n=16

the three numbers are 16, 17 and 18.

2) x+15=4x

3x=15

x=5

radha is now 5 years old

3) the two digits are x and y.

y=3x

(10x+y)+(10y+x)=88

y=3x

11x+11y=88

y=3x

x+y=8

x+3x=8

x=2

the original number can be either 28 or 82.

4) n+(n+2)+(n+4)=48

3n+6=48

3n=42

n=14

the numbers are 14, 16 and 18.

5) y=x+3

10y+x+10x+y=143

y=x+3

11x+11y=143

y=x+3

x+y=13

x+x+3=13

2x=10

x=5

the original number can be 57 or 75.

hope this !