Answer:
20
Step-by-step explanation:
Answer:
Step-by-step explanation:
In Δ ABC & ΔADC
AC bisect ∠BCD given
∠ACB ≅ ∠ACB AC bisect ∠BCD
DC ⊥ AD Given
BC ⊥ AB Given
∠ABC = ∠ADC Right angles are ≅
AC ≅ AC Common to both triangle ΔABC & ΔADC
ΔABC ≅ ΔADC AAS congruent
BC ≅ DC CPCT
Answer:
S1) False. 2³ is 2 × 2 × 2 which is <u>8, not 6.</u>
S2) False. ( 2x / 3y ^ (2) ) ^ (3) =
( (2x) ^ (3) / ( (3y) ^ (2) ) ^ (3) ) = (2)³ × (x)³ / (3)³ × (y²)³ = 8 × x³ / 27 × y<u>⁶</u> = 8x³ / 27y<u>⁶</u><u>.</u>
<u>i.e </u><u>(</u><u>a^</u><u>b</u><u>)</u><u>^</u><u>c</u><u> </u><u>=</u><u> </u><u>a^</u><u>(</u><u>b </u><u>×</u><u> </u><u>c)</u><u>.</u>
S3) True.
S4) True.
S5) False. x^(-3) = 1 / x^(3) ≠ 1 / 3x
