Help me with the second one please
1 answer:
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Answer:
<h2>92 remainder 280</h2>Step-by-step explanation:
So, how much acid is there in 6 gallons? well is 20% acid or (20/100), so the amount of acid in it just (20/100) * 6 or 1.2, the rest is say water.
now, if we want a 90% solution, and say we add "y" gallons, how much acid is in it? well (90/100) * y, or 0.9y.
now let's add "x" gallons of pure acid, now, pure acid is just pure acid, so is 100% acid, how much acid is there in it? (100/100) * x, or 1x or just x.
we know whatever "x" and "y" amounts are, they -> x + 6 = y
and we also know that x + 1.2 = 0.9y
now, if we want a 90% solution, and say we add "y" gallons, how much acid is in it? well (90/100) * y, or 0.9y.
now let's add "x" gallons of pure acid, now, pure acid is just pure acid, so is 100% acid, how much acid is there in it? (100/100) * x, or 1x or just x.
we know whatever "x" and "y" amounts are, they -> x + 6 = y
and we also know that x + 1.2 = 0.9y
Answer:
x ≈ 20.42, y ≈ 11.71
Step-by-step explanation:
Using the cosine ratio on the right triangle on the right, that is
cos20° = =
Multiply both sides by y
y × cos20° = 11 ( divide both sides by cos20° )
y = ≈ 11.71
Using the sine ratio on the right triangle on the left, that is
sin35° = =
=
Multiply both sides by x
x × sin35° = 11.71 ( divide both sides by sin35° )
x = ≈ 20.42