Answer:
<u>Given:</u>
- DC ║ AB
- CM = MB as M is midpoint of BC
i) <u>Since DN and BC are transversals, we have:</u>
- ∠DCM ≅ ∠NBM and
- ∠CDM ≅ ∠BNM as alternate interior angles
<u>As two angles and one side is congruent, the triangles are also congruent:</u>
- ΔDCM ≅ ΔNBM (according to AAC postulate)
So their areas are same.
ii)
<u>The quadrilateral has area of:</u>
- A(ADCB) = A(ADMB) + A(DCM)
<u>And the triangle has area of:</u>
- A(ADN) = A(ADMB) + A(NBM)
Since the areas of triangles DCM and NBM are same, the quadrilateral ADCB has same area as triangle ADN.
Your original equation has no y-variable. It results in x = -2/3 when you solve for x which is a vertical line. Thus, if you want a line parallel to this that goes through point (- 1, - 4), you just set x equal to the x-value in that ordered pair.
Your parallel line is x = - 1.
Answer:
24$
Step-by-step explanation:
its just 9$ + 15$ = 24$
The student added 9 1/2 instead of adding -9 1/2
