Answer:
Step-by-step explanation:
The picture of the question in the attached figure
we know that
In the right triangle of the figure
----> by SOH (opposite side divided by the hypotenuse)
using a calculator
Through any two points there is exactly one?
1 answer:
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Answer:$84.11
Step-by-step explanation:
For
y = (x^2 -4)/((x +2)(x^2 -49))
the numerator factors to (x -2)(x +2), so the factor of (x +2) will cancel with that in the denominator, leaving
y = (x -2)/(x^2 -49)
There are points of discontinuity at the hole, x=-2, and at each of the vertical asymptotes, at x=-7, +7.
The horizontal asymptote is y=0.
y = (x^2 -4)/((x +2)(x^2 -49))
the numerator factors to (x -2)(x +2), so the factor of (x +2) will cancel with that in the denominator, leaving
y = (x -2)/(x^2 -49)
There are points of discontinuity at the hole, x=-2, and at each of the vertical asymptotes, at x=-7, +7.
The horizontal asymptote is y=0.
Step-by-step explanation:
We have been given that AE=BE and .
We can see that angle CEA is vertical angle of angle DEB, therefore, as vertical angles are congruent.
We can see in triangles CEA and DEF that two angles and included sides are congruent.
or
Therefore, by ASA postulate.
Since corresponding parts of congruent triangles are congruent, therefore CE must be congruent to DE.