Please help, legit need to study for so many exams and this is the last thing on my mind to do. Appreciate anybody willing to he
lp <33
1 answer:
You might be interested in
Let's rewrite the system of equations:
y = three over four times x squared minus 3
y = (3/4) x ^ 2 - 3
y = negative three over four times x squared plus 3
y = (-3/4) x ^ 2 + 3
The system of equations is:
y = (3/4) x ^ 2 - 3
y = (-3/4) x ^ 2 + 3
The solutions are:
(-2, 0)
(2, 0)
Answer:
See attached image
y = three over four times x squared minus 3
y = (3/4) x ^ 2 - 3
y = negative three over four times x squared plus 3
y = (-3/4) x ^ 2 + 3
The system of equations is:
y = (3/4) x ^ 2 - 3
y = (-3/4) x ^ 2 + 3
The solutions are:
(-2, 0)
(2, 0)
Answer:
See attached image