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cupoosta [38]
2 years ago
14

Answer Please............​

Mathematics
1 answer:
erik [133]2 years ago
6 0
90 is the right answer
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Prices are going up in the ratio 6:5 Write a 60% decrease as a ratio.
Veronika [31]

Answer:

4 better explanation inbox

Step-by-step explanation:

5 0
2 years ago
)Maxine spent 15 hours doing her homework last week. This week she spent 18 hours doing homework. She says that she spent 120% m
Vlad1618 [11]

Answer:

No she isnt

Step-by-step explanation:

12.5% is for a hour, so if Maxine did 3 hours more than that would be (12.5x 3 ) 37.5%

4 0
3 years ago
Jackson is paid $250.for 5<br> days of Work how much IS<br> he paid per day
Svet_ta [14]
$50 per day.
First you divide 250 dollars by 5 then you will get 50 dollars
5 0
2 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!

7 0
3 years ago
Read 2 more answers
Show that for 5&lt;t&lt;7, the acceleration is given by a=-5t+30
vredina [299]
I hope this helps you



a= -5t+30


5t=30-a


t=30-a/5


5 <30-a/5 <7


5.5 <30-a <7.5


25 <30-a <35


25-30 <30-a-30 <35-30


-5 < -a <5


5>a>-5
8 0
3 years ago
Read 2 more answers
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