Answer:
19.
log9(5x^2 + 10) - log9(10) = 1
<=> log9((5x^2 + 10)/10) = log9(9)
<=> (5x^2 + 10)/10 = 9
<=> 5x^2 + 10 = 90
<=> 5x^2 = 80
<=> x^2 = 16
<=> x = +/- (4)
20.
log5(2x^2 + 4) + log5(3) = 2
<=> log5((2x^2 + 4) x 3) = log3(9)
<=> 6x^2 + 12 = 9
<=> 6x^2 = -3
=> No real x satisfies. ( x^2 always larger or equal to 0)
21.
log6(8) + log6(7 - 2x^2) = 2
<=> log6(8 x (7 - 2x^2)) = log6(36)
<=> 56 - 16x^2 = 36
<=> 16x^2 = 20
<=> x^2 = 5/4
<=> x = +/- sqrt(5/4)
Hope this helps!
:)
Answer: Parallel
i did this yesterday
Answer:
Step-by-step explanation:
y = mx + b.y = 15/13x − 10/13
Both cosθ and sinθ are negative so θ is in the third quadrant ( tan<span>θ is positive)
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tan<span>θ = 5/12
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