2x^2 + 3xy - 4y^2 when x = 2 and y = -4
= 2(2)^2 + 3(2)(-4) - 4(-4)^2 = 2(4) + (-24) - 4(16) = 8 - 24 - 64 = -80
Answer:
27:3
Step-by-step explanation:
multiply the 9 by 3 to get the first part
The equation of the line in its generic form is:
y = mx + b
Where,
m = (y2-y1) / (x2-x1)
For (-1, 3) and (0, 1):
We look for the value of m:
m = (1-3) / (0 - (- 1))
m = (- 2) / (0 + 1)
m = -2
We look for the value of b:
1 = m (0) + b
b = 1
The line is:
y = -2x + 1
For (1, 4) and (0, 2):
We look for the value of m:
m = (2-4) / (0-1)
m = (- 2) / (- 1)
m = 2
We look for the value of b:
2 = m (0) + b
b = 2
The line is:
y = 2x + 2
The system of equations is:
y = -2x + 1
y = 2x + 2
Answer:
the system has one solution
Answer:
x= 51
Step-by-step explanation:
-17=x/-3
(-3)-17=(x/-3)(-3)
51=x
Multiply the denominator of the fraction on both sides of the equation to isolate the variable.
