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3 years ago

Which describes the difference between the two sequences?

Mathematics

1 answer:

mash [69]3 years ago

8 0

Answer:

its a)the first sequence is geometric bc of the common ratio of 2. the second sequence is arithmetic bc of the common difference of 7.

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Mr Nersin likes to ride his bike. He rode for 1/2 hour. After a short rest, he rode 1/10 hour more. He traveled 4 miles in all.

vodomira [7]

The distance traveled by Mr.Nersin = 4 miles
The time for which Mr.Nersin rode first = 1/2 hour
The time for which Mr.Nersin rode after a short rest = 1/10 hour
Then
The total time for which Mr.Nersin rode = (1/2) + (1/10) hour
= (5 + 1)/10 hour
= 6/10 hour
= 3/5 hour
So
Average speed of Mr. Nersin = Distance traveled / Time taken
= 4/(3/5) miles/hour
= (4 * 5)/3 miles/hour
= 20/3 miles/hour
= 6 2/3 miles/hour
= 6.67 miles per hour
So the average speed at which Mr.Nersin traveled is 6 2/3 or 6.67 miles per hour.

6 0

3 years ago

Read 2 more answers

Pleasse help me with this

djverab [1.8K]

Answer:

Step-by-step explanation:

30% off of $18.00

so that makes the price $12.60

7 0

3 years ago

Need this very badly

NeTakaya

((5x5)/2)x2=50 square ft
(8x5)x2= 80 square ft
8x7= 56 square ft
50+80+56=186 square ft
D) 186 square ft

4 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

Write a ratio to express the number of people who preferred WANR to those who preferred WCLM.

iragen [17]

11:12 is the answer to your question

3 0

3 years ago