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3 years ago

Csc2theta=csctheta/2costhetaCan you help verify the identity

Mathematics

1 answer:

Yakvenalex [24]3 years ago

7 0

csc(2x) = csc(x)/(2cos(x))

1/(sin(2x)) = csc(x)/(2cos(x))

1/(2*sin(x)*cos(x)) = csc(x)/(2cos(x))

(1/sin(x))*1/(2*cos(x)) = csc(x)/(2cos(x))

csc(x)*1/(2*cos(x)) = csc(x)/(2cos(x))

csc(x)/(2*cos(x)) = csc(x)/(2cos(x))

The identity is confirmed. Notice how I only altered the left hand side (LHS) keeping the right hand side (RHS) the same each time.

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3 years ago

What is the answer to #30 and #31

svetoff [14.1K]

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30) A) 0 / 15 = 0

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Step-by-step explanation:

30)

0 / 15 = 0

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Answer:

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3 years ago

A filter filled with liquid is in the shape of a vertex-down cone with a height of 9 inches and a diameter of 6 inches at its op

Alina [70]

Answer: Level of the liquid dropping at 28.28 inch/second when the liquid is 2 inches deep.

Step-by-step explanation:

Since we have given that

Height = 9 inches

Diameter = 6 inches

Radius = 3 inches

So, $\dfrac{r}{h}=\dfrac{3}{9}=\dfrac{1}{3}\\\\r=\dfrac{1}{3}h$

Volume of cone is given by

$V=\dfrac{1}{3}\pi r^2h\\\\V=\dfrac{1}{3}\pi \dfrac{1}{9}h^2\times h\\\\V=\dfrac{1}{27}\pi h^3$

By differentiating with respect to time t, we get that

$\dfrac{dv}{dt}=\dfrac{1}{27}\pi \times 3\times h^2\dfrac{dh}{dt}=\dfrac{1}{9}\pi h^2\dfrac{dh}{dt}$

Now, the liquid drips out the bottom of the filter at the constant rate of 4 cubic inches per second, ie $\dfrac{dv}{dt}=-4\ in^3$

and h = 2 inches deep.

$-4=\dfrac{1}{9}\times \pi\times (2)^2\dfrac{dh}{dt}\\\\-9\pi =\dfrac{dh}{dt}\\\\-28.28=\dfrac{dh}{dt}$

Hence, level of the liquid dropping at 28.28 inch/second when the liquid is 2 inches deep.

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2 years ago