We need to assign a value for x to check the possible values of y.
1st inequality: y < -0.75x
X = - 1 ; y < -0.75(-1) ; y < 0.75 possible coordinate (-1,0.75) LOCATED AT THE 2ND QUADRANT
X = 0 ; y < -0.75(0) ; y < 0 possible coordinate (0,0) ORIGIN
X = 1 ; y < -0.75(1) ; y < -0.75 possible coordinate (1,-0.75) LOCATED AT THE 4TH QUADRANT
2nd inequality: y < 3x -2
X = -1 ; y < 3(-1) – 2 ; y < -5 possible coordinate (-1,-5) LOCATED AT THE 4TH QUADRANT
X = 0 ; y < 3(0) – 2 ; y < -2 possible coordinate (0,-2) LOCATED AT THE 4TH QUADRANT
X = 1 ; y < 3(1) – 2 ; y <<span> 1 possible coordinate (1,1) LOCATED AT THE 1ST QUADRANT
The actual solution to the system lies on the 4TH QUADRANT.
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Answer:
its would be a and d
Step-by-step explanation:
6 times 5
Answer:
Whats up good luck... sorry my answer didnt submit
Step-by-stp explanation:
Answer:
2x+ 5
Step-by-step explanation:
