Answer:
y ≥ x^2 - 1
Step-by-step explanation:
First, we can see that the shaded region is above what seems to be a parabola, and we also can see that the lines of the parabola are solid lines (which means that the points on the curve itself are solutions, so the symbol ≥ is used)
Then:
y ≥ a*x^2 + b*x + c
where a*x^2 + b*x + c is the general quadratic equation.
Now let's find the equation for the parabola:
f(x) = a*x^2 + b*x + c
We also can see that the vertex of the parabola is at the point (0, -1)
This means that:
f(0) = -1 = a*0^2 + b*0 + c
= -1 = c
then we have that c = -1
Then:
f(x) = a*x^2 + b*x - 1
Now we can look at the graph again, to see that the zeros of the parabola are at +1 and -1
Which means that:
f(1) = 0 = a*1^2 + b*1 - 1 = a + b - 1
f(-1) = 0 = a*(-1)^2 + b*(-1) - 1 = a - b - 1
Then we got two equations:
a + b - 1 = 0
a - b - 1 = 0
from this we can conclude that b must be zero.
Then:
b = 0
and these equations become:
a - 1 = 0
a - 1 = 0
solving for a, we get:
a = 1
Then the quadratic equation is:
f(x) = 1*x^2 + 0*x - 1
f(x) = x^2 - 1
And the inequality is:
y ≥ x^2 - 1
Hello,
Let's assume a the hypotenuse
b/a=sin 30°==>b=a/2
c/a=cos 30°==>c=√3 / 2 *a
(√3/2 a, a/2,a) are the sides of the triangle.
Answer:
depands
Step-by-step explanation:
on how much ppl
The answer is just -x/4 because you can divide a variable
