Give one positive and one negative coterminal angle for 135°
1 answer:
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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
A)
b)
since QR=QP, that means that QO is an angle bisector, and thus the segments it makes at the bottom of RO and OP, are also equal, thus RO=OP
thus, since the point P is 0.5 units away from the 0, point R is also 0.5 units away from 0 as well, however, is on the negative side, thus R (-0.5, 0)
c)
what's the equation of a line that passes through the points (-0.5, 0) and (0,2)?
b)
since QR=QP, that means that QO is an angle bisector, and thus the segments it makes at the bottom of RO and OP, are also equal, thus RO=OP
thus, since the point P is 0.5 units away from the 0, point R is also 0.5 units away from 0 as well, however, is on the negative side, thus R (-0.5, 0)
c)
what's the equation of a line that passes through the points (-0.5, 0) and (0,2)?