SIMPLIFY (i)35 + 25 x 72 – 51 ÷(10 + 7) (ii)-6 x 9 + 7 – 12 ÷ 3 – 5 Pls give me the correct answer
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Answer:
see explanation
Step-by-step explanation:
(4)
consider the left side
factor the numerator
cosx - cos³x = cosx(1 - cos²x)
cancel sinx on numerator/denominator
= cosxsinx =right side ⇒ verified
(5)
Consider the left side
expand the factors
(1 + cotΘ)² + (1 - cotΘ)²
= 1 + 2cotΘ + cot²Θ + 1 - 2cotΘ + cot²Θ
= 2 + 2cot²Θ
= 2(1 + cot²Θ) ← 1 + cot²Θ = cosec²Θ
= 2cosec²Θ = right side ⇒ verified
(6)
Consider the left side
the denominator simplifies to
cosxtanx = cosx × = sinx
= sinx( +
)
= +
= tanx + 1 = right side ⇒ verified
At the start, the tank contains
(0.02 g/L) * (1000 L) = 20 g
of chlorine. Let <em>c</em> (<em>t</em> ) denote the amount of chlorine (in grams) in the tank at time <em>t </em>.
Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of
(<em>c</em> (<em>t</em> )/(1000 + (10 - 25)<em>t</em> ) g/L) * (25 L/s) = 5<em>c</em> (<em>t</em> ) /(200 - 3<em>t</em> ) g/s
In case it's unclear why this is the case:
The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after <em>t</em> s there will be (1000 + 10<em>t</em> ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time <em>t</em> is (1000 - 15<em>t </em>) L. Then the concentration of chlorine per unit volume is <em>c</em> (<em>t</em> ) divided by this volume.
So the amount of chlorine in the tank changes according to
which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5<em>t</em> )^(5/3), then integrate both sides to solve for <em>c</em> (<em>t</em> ):
There are 20 g of chlorine at the start, so <em>c</em> (0) = 20. Use this to solve for <em>C</em> :