Two squares have sides 12cm and 4cm respectively. What is the ratio of their sides
1 answer:
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Answer:
see the explanation
Step-by-step explanation:
we know that
A shape with two opposite angles equal to 105° could be a quadrilateral, a parallelogram, a rhombus or a trapezoid
Because
<em>A quadrilateral</em>: A quadrilateral is a four-sided polygon. The sum of the interior angles in any quadrilateral must be equal to 360 degrees
so
If the quadrilateral have two opposite angles equal to 105°, then the sum of the other two interior angles must be equal to
<em>A parallelogram</em>: A Parallelogram is a flat shape with opposite sides parallel and equal in length. Opposite angles are congruent and consecutive angles are supplementary
so
If the parallelogram have two opposite angles equal to 105°, then the measure of each of the other two congruent interior angles must be equal to
<em>A rhombus</em>: A Rhombus is a flat shape with 4 equal straight sides. A rhombus looks like a diamond. All sides have equal length. Opposite sides are parallel. Opposite angles are congruent and consecutive angles are supplementary
so
If the Rhombus have two opposite angles equal to 105°, then the measure of each of the other two congruent interior angles must be equal to
<em>A trapezoid</em>: A trapezoid is a 4-sided flat shape with straight sides that has a pair of opposite sides parallel
so
If the trapezoid have two opposite angles equal to 105°, then the sum of the other two interior angles must be equal to
By definition of absolute value, you have
or more simply,
On their own, each piece is differentiable over their respective domains, except at the point where they split off.
For <em>x</em> > -1, we have
(<em>x</em> + 1)<em>'</em> = 1
while for <em>x</em> < -1,
(-<em>x</em> - 1)<em>'</em> = -1
More concisely,
Note the strict inequalities in the definition of <em>f '(x)</em>.
In order for <em>f(x)</em> to be differentiable at <em>x</em> = -1, the derivative <em>f '(x)</em> must be continuous at <em>x</em> = -1. But this is not the case, because the limits from either side of <em>x</em> = -1 for the derivative do not match:
All this to say that <em>f(x)</em> is differentiable everywhere on its domain, <em>except</em> at the point <em>x</em> = -1.