Answer:
Step-by-step explanation: 40
Answer:
Sum of the first 15 terms = -405
Step-by-step explanation:
a + 3d = -15 (1)
a + 8d = -30 (2)
Where,
a = first term
d = common difference
n = number of terms
Subtract (1) from (1)
8d - 3d = -30 - (-15)
5d = -30 + 15
5d = -15
d = -15/5
= -3
d = -3
Substitute d = -3 into (1)
a + 3d = -15
a + 3(-3) = -15
a - 9 = -15
a = -15 + 9
a = -6
Sum of the first 15 terms
S = n/2[2a + (n − 1) × d]
= 15/2 {2×-6 + (15-1)-3}
= 7.5{-12 + (14)-3}
= 7.5{ -12 - 42}
= 7.5{-54}
= -405
Sum of the first 15 terms = -405
The halves of the rhombus separated by line BD are equilateral triangles.
∠BAD = 60°
so
∠ADM is 0.80*60° = 48°.
∠BMD is an exterior angle to ΔAMD, so is equal to the sum
∠BMD = ∠BAD + ∠ADM
= 60° +48° = 108°
∠BMD = 108°
Easy all you do is <span>14<span>(6/1)</span></span><span>=<span><span>(14/1)</span><span>(6/1)</span></span></span><span>=<span><span>(14)(6)</span><span>(1)(1)</span></span></span><span>=84</span>