Answer:
B C and D
Step-by-step explanation:
Mark brainliest
P.S this was epert verified
Answer:
Δ BEC ≅ Δ AED
Step-by-step explanation:
Consider triangles BCA and ADB. Each of them share a common side, AB. Respectively each we should be able to tell that AD is congruent to BC, and DB is congruent to CA, so by SSS the triangles should be congruent.
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So another possibility is triangles BEC, and AED. As you can see, by the Vertical Angles Theorem m∠BEC = m∠ADE, resulting in the congruency of an angle, rather a side. As mentioned before AD is congruent to BC, and perhaps another side is congruent to another in the same triangle. It should be then, by SSA that the triangles are congruent - but that is not an option. SSA does is one of the exceptions, a rule that is not permitted to make the triangles congruent. Therefore, it is highly unlikely that triangles BEC and AED are congruent, but that is what our solution, comparative to the rest.
Δ BEC ≅ Δ AED .... this is our solution
Answer:
x = - 2.5
Step-by-step explanation:
Given that the sketch represents
y = x² + bx + c
The graph crosses the y- axis at (0 , - 14), thus c = - 14
y = x² + bx - 14
Given the graph crosses the x- axis at (2, 0), then
0 = 2² + 2b - 14
0 = 4 + 2b - 14 = 2b - 10 ( add 10 to both sides )
10 = 2b ( divide both sides by 2 )
b = 5
y = x² + 5x - 14 ← represents the graph
let y = 0 , then
x² + 5x - 14 = 0 ← in standard form
(x + 7)(x - 2) = 0 ← in factored form
Equate each factor to zero and solve for x
x + 7 = 0 ⇒ x = - 7
x - 2 = 0 ⇒ x = 2
The x- intercepts are x = - 7 and x = 2
The vertex lies on the axis of symmetry which is midway between the x- intercepts, thus
the x- coordinate of the turning point is
=
= - 2.5