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nika2105 [10]
3 years ago
8

A jet flew to the maintenance facility and back. The trip there took ten hours and the trip back took four hours. It averaged 42

0 mph on the return trip. Find the average speed of the trip there.
Mathematics
1 answer:
Lana71 [14]3 years ago
3 0
Speed is distance over time

miles per hour is distance over time

so if it averaged 420 miles an hour

is it not 420 miles per hour?
 
perhaps i am misunderstanding something?


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Three consecutive integers are n, n+1, and n+2.
Vera_Pavlovna [14]
<span>We convert the sentences to mathematical expressions as follows


"</span>Four times the sum of the least and greatest integers" : 4[n+(n+2)]

"is 12 less than three times the least integer": = 3n-12



So we have:

4[n+(n+2)]=3n-12   (this is the equation)

4[2n+2]=3n-12
8n+8=3n-12

8n-3n=-12-8
5n=-20

n=-20/5=-4


Answer: the equation is 4[n+(n+2)]=3n-12 , the least integer is -4
8 0
3 years ago
Given the inverse or rational function <br> Find the value of when =−3
Sveta_85 [38]

Answer:

inverse= 3. rational=-3/1

This is your answer

5 0
3 years ago
Determine whether the point (-3,-6) is in the solution set of the system of inequalities below
Sergio039 [100]

Answer:

A

Step-by-step explanation:

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7 0
3 years ago
Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab
Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx

If f(x)=x^2, then

f'(x)=2x

and

b=0\\ \\a=2

Therefore,

SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx

Apply substitution

x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du

Then

SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du

Now

\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))

Hence,

SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}

3 0
3 years ago
I need to know what is 5/8 x 4??
Aloiza [94]

Answer:

\frac{20}{8}=\frac{5}{2}= 2\frac{1}{2}

Step-by-step explanation:

\frac{5}{8}*\frac{4}{1}=\frac{20}{8}=2\frac{1}{2}

4 0
2 years ago
Read 2 more answers
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