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
Answer:
The answer is
X=1
Y=1
Step-by-step explanation:
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Answer:
95
students in the seventh grade have taken swimming lessons.
Explanation:
We calculate
38
%
of
250
using the following formula:
x
=
250
×
38
100
x
=
25
0
×
38
10
0
(cancel by 10)
x
=
5
25
×
38
2
10
(cancel by 5)
x
=
5
×
19
38
1
2
(cancel by 2)
x
=
5
×
19
x
=
95
Step-by-step explanation:
Answer:
if there is gonna be 100 cookies an 400 raisins in the batch there should be a 1% chance that no 1 or more cookie will come out with out raisins. You have more raisins than the number of cookies which if u ever bake cookies when adding in things like that you should add in as much raisin as there is cookie.
Step-by-step explanation:
One of the properties of a parallelogram is that its opposite sides are parallel and congruent.
Segment AB and CD are opposite sides of the parallelogram and is therefore, congruent.
Therefore, the reason for CD≅ AB is: "Opposite sides of a parallelogram/rhombus/rectangle/square are congruent."
For the next statement, since CD≅AB and AB≅CE, then by Transitive Property, CD≅CE.
Since CD and CE are sides of a triangle and are congruent as stated in Statement 3, then ∠E ≅ ∠CDE because in a triangle, angles opposite of congruent sides are congruent.
In addition, we can say that ∠A ≅ ∠CDE because parallel lines (AB and CD) cut by a transversal (AE) form congruent corresponding angles.
Lastly, since ∠A ≅ ∠CDE and ∠CDE ≅ ∠E, we can say that ∠A ≅ ∠E by Transitive Property.