6+ 4×whatever y equals +9
Given:
In circle O, m∠R = 30.8°.
To find:
The m∠NOQ
Solution:
Central angle theorem: According to this theorem, the central angle is always twice of subtended angle on the same arc.
Angle NOQ and angle NRQ are on the same arc but ∠NOQ is the central angle and ∠NRQ is the subtended angle on the arc NQ.
Using central angle theorem, we get
Therefore, .
Answer:
Step-by-step explanation:
2√54+5√24
2 * sqrt(9*6) + 5*sqrt(4*6)
2*sqrt(9) *sqrt(6) + 5*sqrt(4) * sqrt(6)
2*3 sqrt(6) + 5*2 sqrt(6)
6sqrt(6)+10sqrt(6)
16sqrt(6)
Answer:
x = 136/35; y = -⁹/₁₀
Step-by-step explanation:
(1) 7x + 8y = 20
(2) 7x – 2y = 29 Subtract (2) from (1)
10y = -9 Divide each side by 10
(3) y = -⁹/₁₀ Substitute (3) into (1)
7x - 2(-⁹/₁₀) = 29
7x + 18/10 = 29 Subtract 18/10 from each side
7x = 29 - 18/10
7x = (290 - 18)/10
7x = 272/10 Divide each side by 7
x = 272/70
x = 136/35
x = 136/35; y = -⁹/₁₀
Check:
(1) 7(136/35) + 8(-⁹/₁₀) = 20
136/5 - 72/10 = 20
136/5 - 36/5 = 20
100/5 = 20
20 = 20
(2) 7(136/35) – 2(-⁹/₁₀) = 29
136/5 + 18/10 = 29
136/5 + ⁹/₅ = 29
145/5 = 29
29 = 29
