For each parabola, you have to do a system of equation with the line y = x - 5 and find which of them has one real solution.
For that you can solve the systems and find the roots, but you can also use the rule that to have one real solution the discriminant of the quadratiic equation (b^2 - 4ac) has to be zero.
1)
y = x - 5
y = x^2 + x - 4
=> x - 5 = x^2 + x - 4
=> x^2 + 1 = 0
It is easy to tell, by simple inspection, that this equation has not real solutions.
2)
y = x -5
y = x^2 + 2x - 1
=> x - 5 = x^2 + 2x - 1
=> x^2 + x + 4 = 0
discriminant = b^2 - 4ac = 1^2 - 4(1)(4) = 1 - 16 = - 15
A negative discriminant means that there are not real solutions.
3)
y = x - 5
y = x^2 + 6x + 9
x - 5 = x^2 + 6x + 9
=> x^2 + 5x + 14 = 0
=> b^2 - 4ac = 5^2 - 4(1)(14) = 25 - 56 = - 31 => no real solutions
4)
y = x - 5
y = x^2 + 7x + 4
x - 5 = x^2 + 7x + 4
=> x^2 + 6x + 9 = 0
=> b^2 - 4ac = 6^2 - 4(1)(9) = 36 - 36 = 0 => the system has one real solution.
By the way, that solution is easy to find because you can factor the equation as: (x + 3)^2 = 0 => x = - 3
I can’t see make clearer plz
The formula for area of a parallelogram is a=b•h
a=42•28=1,176
A = L * W
A = 4 3/4 * 3 2/5.....turn them into improper fractions
A = 19/4 * 17/5
A = 323/20 or 16 3/20 yds^2 <==
Answer:
The answer is 1
Step-by-step explanation:
5w² – 5w = 0
Add 5w both side
5w² – 5w + 5w = 5w – 5w
5w² = 5w
Now, Divide both side by 5w we get,
5w²/5w = 5w/5w
w = 1
Thus, The value of w is 1