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Rom4ik [11]
3 years ago
6

Variable c is 5 more than variable

Mathematics
2 answers:
olga_2 [115]3 years ago
8 0
More means to add NOT multiply
less means subtract therefore,

c= a+5
c= a-3
is the the correct choice

hope this helps
shutvik [7]3 years ago
7 0
First, let us restate the given conditions

c is 5 more than variable a ( c = a + 5)
c is also three less than variable a (c = a - 3)

Now, lets look at the answer choices and or given
c = a − 5 
c = a + 3
Here, c is 5 less than "a"...so automatically disqualified

a = c + 5
a = 3c − 3 
Here, we have to get "C" by itself in both top and bottom equation.
So,
Simplified version :
c = a - 5
Here, c is 5 less than "a"...so automatically disqualified


a = c − 5
a = 3c + 3
Here also, we have to get "C" by itself in both top and bottom equation.
So, 
simplified version:
c = a + 5
Here, c is 5 more than "a"...so we continue
c = (a - 3) / 3
Here, c is 3 less than "a" <u>divided by 3</u><u /> . So, this is not correct

c = a + 5
c = a − 3
Here, c is 5 more than "A"
Also, c is 3 less than "a"
Which satisfies the given.

 So, our answer is going to be the last one:
 
 c = a + 5
 c = a - 3
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Answer:

The position vector of point C is <-3 , -17 , 8> or -3i - 17j + 8k

Step-by-step explanation:

* Lets revise how to solve the problem

- If the endpoints of a segment are (x1 , y1 , z1) and (x2 , y2 , z2), and

 point (x , y , z) divides the segment externally at ratio m1 :m2, then

 x=\frac{m_{1}x_{2}-m_{2}x_{1}}{m_{1} -m_{2}},y=\frac{m_{1}y_{2}-m_{2}y_{1}}{m_{1}-m_{2}},z=\frac{m_{1}z_{2}-m_{2}z_{1}}{m_{1}-m_{2}}

* Lets solve the problem

∵ AB is a segment where A = (3 , 1 , 2) and B = (1 , - 5 , 4)

∵ Point C lies on line AB such that AC : BC=3 : 2

∵ From the ratio AC = 3/2 AB

∴ C divides AB externally

- Lets use the rule above to find the coordinates of C

- Let Point A is (x1 , y1 , z1) , point B is (x2 , y2 , z2) and point C is (x , y , z)

 and AC : AB is m1 : m2

∴ x1 = 3 , x2 = 1

∴ y1 = 1 , y2 = -5

∴ z1 = 2 , z2 = 4

∴ m1 = 3 , m2 = 2

- By using the rule above

∴ x=\frac{3(1)-2(3)}{3-2}=\frac{3-6}{1}=\frac{-3}{1}=-3

∴ y=\frac{3(-5)-2(1)}{3-2}=\frac{x=-15-2}{1}=\frac{-17}{1}=-17

∴ z=\frac{3(4)-2(2)}{3-2}=\frac{12-4}{1}=\frac{8}{1}=8

∴ The coordinates fo point c are (-3 , -17 , 8)

* The position vector of point C is <-3 , -17 , 8> or -3i - 17j + 8k

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