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balandron [24]
3 years ago
14

Which quadrant contains the point named by (2,-5)?

Mathematics
2 answers:
valentina_108 [34]3 years ago
7 0
No idea hope you figure it out tho !!
maybe 3?
uranmaximum [27]3 years ago
5 0

Answer:

Quadrant IV

Step-by-step explanation:

(+,+)= Quadrant I

(-,+)= Quadrant II

(-,-)= Quadrant III

(+,-)= Quadrant IV

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gulaghasi [49]
-(-z + 3) equals z - 3
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z = 8
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11

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The sum of 10 and twice Mai’s age
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Can I get the full question please

Step-by-step explanation:

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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:

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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:

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Distribute the 2:

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Substitute back secant for u:

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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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