The solution to the problem is as follows:
If you use Ln you can get rid of e.
ln(e^0.4x) = ln(0.4)
0.4x = ln(0.4)
x = (ln(0.4))/0.4
<span>x = -2.29
</span>
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The answer is 3/5
Just keep the distributive property in mind.
F - g(2) = 3x^2 + 1 - ( 1 - x )
= 3x^2 + 1 - 1 + x
= 3x^2 + x
= 3(2)^2 + 2
= 14
the answer is B.14
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tan2x*cotx - 3 = 0
We know that: tan2x = sin2x/cos2x and cotx = cosx/sinx
==> sin2x/cos2x *cosx/sinx = 3
Now we know that sin2x = 2sinx*cosx
==> 2sinxcosx/cos2x * cosx/sinx = 3
Reduce sinx:
==> 2cos^2 x/ cos2x = 3
Now we know that cos2x = 2cos^2 x-1
==> 2cos^2 x/(2cos^2 x -1) = 3
==> 2cos^2 x = 3(2cos^2 x -1)
==> 2cos^2 x = 6cos^2 x - 3
==> -4cos^2 x= -3
==> 4cos^2 x = 3
==> cos^2 x = 3/4
==> cosx = +-sqrt3/ 2
<span>==> x = pi/6, 5pi/6, 7pi/6, and 11pi/6</span>
No. he used 50% on the rides, and we have to convert the fraction to a percent by dividing 4&1 which equals 0.25, then we multiply that by 100 to get the percentage he used on video games which equals 25%. 50+25= 75% used tickets for both rides and video games. So then we subtract 75 from 100 and get 25% which means he used 25% of the tickets on batting cages. Your answer is no, he used 25% of his tickets on the batting cages.
