I think it's D. I will admit that I'm an educated guess.
<span>The value of y when x=3 when the value of y varies directly with x^2, and y = 150 when x = 5 is 54. Since the value of y varies directly with x2, then: y = k * x^2, where k is constant. When y = 150 and x = 5, the value of constant is: 150 = k * 5^2. 150 = k * 25. k = 150/25. k = 6. Thus, when x = 3, the value of y will be: y = 6 * 3^2. y = 6 * 9. y = 54.</span>
Answer:
Substitution method can be used
Step-by-step explanation:
Given the system of equations
y = 2x-1 ....1
-12x + 3y = 9 ....2
The best method to use id the substitution method
Substitute equation 1 into 2;
From 2;
-12x + 3y = 9
-4x + y = 3
-4x + 2x-1 = 3
-2x -1 = 3
-2x = 3+1
-2X = 4
x = -4/2
x = -2
Substitute x = -2 into 1
y = 2x - 1
y = 2(-2)-1
y = -4-1
y = -5
Hence the solution to the system of equation is (-2, -5)
Answer:
Step-by-step explanation:
Given expression,
=>
We see that 
is common in both numerator and denominator
hence canceling out we get:-
Answer=
