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3 years ago

322 divided by 14 using quotients

Mathematics

1 answer:

vodka [1.7K]3 years ago

6 0

322 divided by 14 equals 23

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The graph below shows Kate's distance from her home (y), in miles, after a certain amount of time (x), in minutes:

Aliun [14]

The correct answer for this question is this one: "Tim."
Tim said that "<span><em>She drives at a variable speed for 2 minutes, then drives at a constant speed for 4 minutes, and finally stops at the parking area for the next 2 minutes." </em></span>Hope this helps answer your question and have a nice day ahead.

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3 years ago

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5(2x-5) = 1/2(18x+40)

andrew11 [14]

Answer:

x = 45

Step-by-step explanation:

5 (2x - 5) = 1/2 (18x + 40)

10x - 25 = 9x + 20

10x = 9x + 45

x = 45

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3 years ago

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A colony of 500 bacteria doubles in size every 200 hours. What will the population be 400 hours from now?

mote1985 [20]

Answer:

1000

Step-by-step explanation

200+200=400

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3 years ago

(8x+2) by (5x+4) what’s the area

ki77a [65]

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Step-by-step explanation:

3 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