Answer:
Step-by-step explanation:
We are given that a and b are rational numbers where
and x is irrational number .
We have to prove a+bx is irrational number by contradiction.
Supposition:let a+bx is a rational number then it can be written in
form
where
where p and q are integers.
Proof:
After dividing p and q by common factor except 1 then we get

r and s are coprime therefore, there is no common factor of r and s except 1.
where r and s are integers.


When we subtract one rational from other rational number then we get again a rational number and we divide one rational by other rational number then we get quotient number which is also rational.
Therefore, the number on the right hand of equal to is rational number but x is a irrational number .A rational number is not equal to an irrational number .Therefore, it is contradict by taking a+bx is a rational number .Hence, a+bx is an irrational number.
Conclusion: a+bx is an irrational number.
(3x+2) (x-7)
FOIL
first 3x*x = 3x^2
outer 3x*-7 = -21x
inner = 2*x = 2x
last 2*-7 = -14
add them together = 3x^2 -21x+2x -14
combine like terms 3x^2 -19x -14
So this is what i did but im not sure if its 100% correct
585 / 9 to get the number of groups = 65
then you take 65 x 3 for number of students in each group and you get the final answer of 195 students
Answer:
(3, -3) is correct
Step-by-step explanation:
-4 × 3 = -12. -12 + 9 = -3
Step-by-step explanation:
the equations are :-
=》d = 40m + 200
=》d = 50m
here, by equalising both equations.
( since, d = d )
=》40m + 200 = 50m
=》50m - 40m = 200
=》10m = 200
=》m = 200 ÷ 10
=》m = 20
now,
=》d = 50m
=》d = 50 × 20
=》d = 1000