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3 years ago

7

Is the radius of a cylinder basically the base or is it the diameter?

2 answers:

elena-s [515]3 years ago

8 0

No, The base of a circle would be the area of the circle or the circle itself (top or bottom) of the cylinder

zalisa [80]3 years ago

7 0

The radius is half of the Diameter. So it would be neather.

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How many ounces are in a pound and a half

faltersainse [42]

16.5 oz are in a pound a half

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3 years ago

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Tyrone has $8.06.How many bagels can he buy if each bagel cost$0.65?

aleksklad [387]

Answer:

12

Step-by-step explanation:

so $8.06 / .65 = 12.4 but you cannot buy .4 of a bagel so the answer would just be 12

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4 years ago

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At noon of a certain day, when 5 pens and 3 pencils were placed in a drawer, the ratio of the number of pens to the number of pe

Molodets [167]

The question is probably incomplete. What i am assuming is that the person might ask for the previous ratio or the total number of pens and pencil in the drawer.

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3 years ago

You received a gift in the mail. The length of the box is 12

lyudmila [28]

Answer:

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Step-by-step explanation:

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3 years ago

Calculus 2 master needed; stuck on evaluating the integral please show steps <img src="https://tex.z-dn.net/?f=%5Cint%20%7Bsec%2

kakasveta [241]

Answer:

$\int \sec(\frac{x}{2})\tan^5({\frac{x}{2}})dx=\frac{2\sec^5(\frac{1}{2}x)}{5}-\frac{4\sec^3(\frac{1}{2}x)}{3}+2\sec(\frac{1}{2}x)+C$

Step-by-step explanation:

So we have the integral:

$\int \sec(\frac{x}{2})\tan^5({\frac{x}{2}})dx$

First, let's use substitution to get rid of the x/2. I'm going to use the variable y. So, let y be x/2. Thus:

$y=\frac{x}{2}\\dy=\frac{1}{2}dx\\2dy=dx$

Therefore, the integral is:

$=2\int \sec(y)\tan^5(y)dy$

Now, as you had done, let's expand the tangent term. However, let's do it to the fourth. Thus:

$=2\int \sec(y)\tan^4(y)\tan(y)dy$

Now, we can use a variation of the trigonometric identity:

$\tan^2(y)+1=\sec^2(y)$

So:

$\tan^2(y)=\sec^2(y)-1$

Substitute this into the integral. Note that tan^4(x) is the same as (tan^2(x))^2. Thus:

$=2\int \sec(y)(\tan^2(y))^2\tan(y)dy\\=2\int \sec(y)(\sec^2(y)-1)^2\tan(y)dy$

Now, we can use substitution. We will use it for sec(x). Recall what the derivative of secant is. Thus:

$u=\sec(y)\\du=\sec(y)\tan(y)dy$

Substitute:

$2\int (\sec^2(y)-1)^2(\sec(y)\tan(y))dy\\=2\int(u^2-1)^2 du$

Expand the binomial:

$=2\int u^4-2u^2+1du$

Spilt the integral:

$=2(\int u^4 du+\int-2u^2du+\int +1du)$

Factor out the constant multiple:

$=2(\int u^4du-2\int u^2du+\int(1)du)$

Reverse Power Rule:

$=2(\frac{u^{4+1}}{4+1}-2(\frac{u^{2+1}}{2+1})+(\frac{u^{0+1}}{0+1}}))$

Simplify:

$=2(\frac{u^5}{5}-\frac{2u^3}{3}+u)$

Distribute the 2:

$=\frac{2u^5}{5}-\frac{4u^3}{3}+2u$

Substitute back secant for u:

$=\frac{2\sec^5(y)}{5}-\frac{4\sec^3(y)}{3}+2\sec(y)$

And substitute back 1/2x for y. Therefore:

$=\frac{2\sec^5(\frac{1}{2}x)}{5}-\frac{4\sec^3(\frac{1}{2}x)}{3}+2\sec(\frac{1}{2}x)$

And, finally, C:

$=\frac{2\sec^5(\frac{1}{2}x)}{5}-\frac{4\sec^3(\frac{1}{2}x)}{3}+2\sec(\frac{1}{2}x)+C$

And we're done!

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3 years ago

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