Answer:
(x, y) = (1, 1/3)
Step-by-step explanation:
The x-coefficient in the first equation is -2 times that in the second equation, so adding twice the second equation to the first will eliminate x:
(4x -9y) +2(-2x +3y) = (1) +2(-1)
-3y = -1 . . . . simplify
y = 1/3 . . . . . divide by -3
The y-coefficient in the first equation is -3 times the y-coefficient in the second equation, so adding 3 times the second equation to the first will eliminate y:
(4x -9y) +3(-2x +3y) = (1) +3(-1)
-2x = -2 . . . . . . simplify
x = 1 . . . . . . . . . divide by -3
The solution is (x, y) = (1, 1/3).
Answer:
480
Step-by-step explanation:
Answer:
I think that the best unit rate is Grande: 16oz ; $4.25
Answer:
Option A is correct
The function 
has real zeroes at x =-10 and x =-6
Explanation:
Given: The real zeroes or roots are x = -10, and x = -6
To find the quadratic function of degree 2.
where α,β are real roots. ....[1]
Here, α= -10 and β= -6
Sum of the roots:
α+β = -10+(-6) = -10-6 = -16
Product of the roots:
αβ = (-10)(-6)= 60
Substitute these value in equation [1] we have;
Therefore, the quadratic function for the real roots at x =-10 and x =-6 ;
Answer:
hole at x=-3
Step-by-step explanation:
The hole is the discontinuity that exists after the fraction reduces. (Still doesn't exist for original of course)
The discontinuities for this expression is when the bottom is 0. x^2-9=0 when x=3 or x=-3 since squaring either and then subtracting 9 would lead to 0.
So anyways we have (x+3)/(x^2-9)
= (x+3)/((x-3)(x+3))
Now this equals 1/(x-3) with a hole at x=-3 since the x+3 factor was "cancelled" from the denominator.
