Answer specify
Step-by-step explanation:
Don’t know
All squares are rectangles
all squares are rhombuses
Answer:
1.) Triangle ABC is congruent to Triangle CDA because of the SAS theorem
2.) Triangle JHG is congruent to Triangle LKH because of the SSS theorem
Step-by-step explanation:
Alright. Let's start with the 1st figure. How do we prove that triangles ABC and CDA (they are named properly) are congruent? First, we can see that segments BC and AD have congruent markings, so that can help us. We also see a parallel marking for those segments as well, meaning that the diagonal AC is also a transversal for those parallel segments. That means we can say that angle CAD is congruent to angle ACB because of the alternate interior angles theorem. Then, the 2 triangles also share the side AC (reflexive property).
So, we have 2 congruent sides and 1 congruent angle for each triangle. And in the way they are listed, this makes the triangles congruent by the SAS theorem since the angle is adjacent to the 2 sides that are congruent.
The second figure is way easier. As you can clearly see by the congruent markings on the diagram, all the sides on one triangle are congruent to the other. So, since there are 3 sides congruent, we can say the triangles JHG and LKH are congruent by the SSS theorem.
Step-by-step explanation:
<em>cotx-</em><em> </em><em>tanx</em><em>=</em><em> </em><em>2cot2x</em>
<em>L.H.S</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>cosx</em><em>/</em><em>sinx-</em><em> </em><em>sinx</em><em>/</em><em>cosx</em><em> </em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>(</em><em>cos</em><em>²</em><em>x-</em><em> </em><em>sin</em><em>²</em><em>x</em><em>)</em><em>/</em><em> </em><em>sinxcosx</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>[</em><em> </em><em>cos</em><em>²</em><em>x-</em><em>(</em><em>1</em><em>-cos</em><em>²</em><em>x</em><em>)</em><em>]</em><em>/</em><em>sinxcosx</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>(</em><em> </em><em>2cos</em><em>²</em><em>x-1</em><em>)</em><em>/</em><em>sinxcosx</em>
<em> </em><em> </em><em> </em><em>2cosx</em><em>/</em><em> </em><em>2sinxcosx</em>
<em> </em><em> </em><em> </em><em>2cot2x</em>
Yes. A is the right graph because the line must travel through two points: (0 ,1)and(-3, 0)
