Let f(x) = x² + 6x²-x+ 5 then ,
number to be added be P
then,
f(x) = x² + 6x²-x+ 5 +P
According to the qn,
(x+3) is exactly divisible by zero then,
R=0
comparing .. we get a= -3
now by remainder theorm
R=f(a)
0=f(-3)
0=(-3)² + 6(-3)²-(-3)+ 5 + P
0= 9 + 54 + 3 + 5 + P
-71=P
therefore, -71 should be added.
Hope you understand
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
Answer:
I think it is 24 cus it's more bigger and cooler and hotter and shorter and butter
Double my tens digit to get my ones digit and double me and i am less than 50?
12,24
Hope I helped, good luck!
