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
Answer:
Step-by-step explanation:
Answer:
x= -3 and y= 0
Step-by-step explanation:
5x+2y=-15
<u>2x-2y=-6 </u>
<u>7x =-21</u>
x= -3
Putting value of x in equation 1
5(-3) +2y=-15
-15+2y= -15
2y= 0
y= 0
This can be solved with the help of matrices
In matrix form the above equations can be written in the form
![\left[\begin{array}{ccc}5&2\\2&-2\/\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%262%5C%5C2%26-2%5C%2F%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}x\\y\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}-15\\-6\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-15%5C%5C-6%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Let
= A = X and = B
Then AX= B
or X= A⁻¹ B
where A⁻¹= adj A/ ║A║ where mod A≠ 0
adj A= ![\left[\begin{array}{ccc}-2&-2\\-2&5\/\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%26-2%5C%5C-2%265%5C%2F%5Cend%7Barray%7D%5Cright%5D)
║A║= ( 5*-2- 2*2)= -10-4= -14≠0
X= A⁻¹ B
=- 1/14
=- 1/14 ![\left[\begin{array}{ccc}-2*-15&+ -2*-6\\-2*-15&+ 5*-6\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%2A-15%26%2B%20-2%2A-6%5C%5C-2%2A-15%26%2B%205%2A-6%5C%5C%5Cend%7Barray%7D%5Cright%5D)
=- 1/14 ![\left[\begin{array}{ccc} 30&+12\\30&+-30\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%2030%26%2B12%5C%5C30%26%2B-30%5C%5C%5Cend%7Barray%7D%5Cright%5D)
=- 1/14 ![\left[\begin{array}{ccc}42\\0\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D42%5C%5C0%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}-42/14\\0/-14\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-42%2F14%5C%5C0%2F-14%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}-3\\0\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%5C%5C0%5C%5C%5Cend%7Barray%7D%5Cright%5D)
From here x= -3 and y= 0
Solution Set = [(-3,0)]
Answer: x=2 and y=-3
Step-by-step explanation: you’re welcome
Answer:
I am pretty sure its y={5}/{4}x-5
Step-by-step explanation:
