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alex41 [277]
3 years ago
5

Use Cavalieri’s Principle to calculate the exact volume of an oblique cylinder with a radius of 5 inches and a height of 16 inch

es.
Mathematics
2 answers:
andrew-mc [135]3 years ago
7 0

Answer:

V=400{\pi} in^3

Step-by-step explanation:

Cavalieri’s Principle: If in two solids of equal altitude, the sections made by planes parallel to and at the same distance from their respective bases are always equal, then the volumes of the two solids are equal.

Thus, we are given that the radius of the cylinder is 5 inches and height is 16 inches, therefore volume of oblique cylinder is given as:

V={\pi}r^2h

Substituting the given values,, we have

V={\pi}(5)^2(16)

V={\pi}(25)(16)

V=400{\pi} in^3

Thus, the volume of the oblique cylinder is 400π cubic inches.

Oliga [24]3 years ago
5 0
V= (pi)r^29(h)
V= (pi) 25(16)

V= 400(pi) in^3
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A. Evaluate ∫20 tan 2x sec^2 2x dx using the substitution u = tan 2x.
irakobra [83]

Answer:

The integral is equal to 5\sec^2(2x)+C for an arbitrary constant C.

Step-by-step explanation:

a) If u=\tan(2x) then du=2\sec^2(2x)dx so the integral becomes \int 20\tan(2x)\sec^2(2x)dx=\int 10\tan(2x) (2\sec^2(2x))dx=\int 10udu=\frac{u^2}{2}+C=10(\int udu)=10(\frac{u^2}{2}+C)=5\tan^2(2x)+C. (the constant of integration is actually 5C, but this doesn't affect the result when taking derivatives, so we still denote it by C)

b) In this case u=\sec(2x) hence du=2\tan(2x)\sec(2x)dx. We rewrite the integral as \int 20\tan(2x)\sec^2(2x)dx=\int 10\sec(2x) (2\tan(2x)\sec(2x))dx=\int 10udu=5\frac{u^2}{2}+C=5\sec^2(2x)+C.

c) We use the trigonometric identity \tan(2x)^2+1=\sec(2x)^2 is part b). The value of the integral is 5\sec^2(2x)+C=5(\tan^2(2x)+1)+C=5\tan^2(2x)+5+C=5\tan^2(2x)+C. which coincides with part a)

Note that we just replaced 5+C by C. This is because we are asked for an indefinite integral. Each value of C defines a unique antiderivative, but we are not interested in specific values of C as this integral is the family of all antiderivatives. Part a) and b) don't coincide for specific values of C (they would if we were working with a definite integral), but they do represent the same family of functions.  

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vitfil [10]
Ie 32^2 - 4ac=0 so ac=256 and a+c=130 
So, that means:2 equations ac=256 a+c=130 ie c=130-a 
So,a(130-a)=256 so 130a -a^2=256 ie a^2-130a +256=0 (a-2)(a-128)=0 
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