4(2x-4)^2 - 3
4(4x^2 - 2(4)(2x) + 16) - 3
4(4x^2 -16x + 16) - 3
16x^2 - 32x + 32 - 3
16x^2 - 32x + 29
Answer:
Step-by-step explanation:
x-2y=-2 x=2y-2
7x-3y=19
substitute x with 2y-2
7(2y-2)-3y=19 14y-14-3y=19 11y=33 y=3
substitue 3 for y
x-2(3)=-2 x-6=-2 x=4
x=4, y=3
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Answer:
Step-by-step explanation:x<-5/2
Answer:
answer is 3
Step-by-step explanation:
add 5 minus 10
Answer:
Step-by-step explanation:
we know that
The solution for region VI is the shaded area above the solid line 
and between the positive values of x and the negative values of y
so
The system of inequalities that represent the region VI is
