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3 years ago

A location in space is the definition of a

Mathematics

1 answer:

elena-14-01-66 [18.8K]3 years ago

4 0

<span>A location in space is the definition of a Point.</span>

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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

The temperature at 7 a.m. was -11°C. The temperature increased by 9°C during the

LiRa [457]

The answer is -6 degrees Celsius

6 0

3 years ago

How many quarter pound buyers can george make out of 3 and 1/2 pounds of ground beef?

weeeeeb [17]

Divide the total amount by the portion size.

= 3 1/2 ÷ 1/4
convert 3 1/2 to improper fraction

= (2*3+1)/2 ÷ 1/4
= 7/2 ÷ 1/4

to divide a fraction, we multiply by the reciprocal/inverse of 1/4

= 7/2 * 4/1
multiply numerators, then multiply denominators

= (7*4)/(2*1)
= 28/2

= 14

ANSWER: 14

Hope this helps! :)

8 0

3 years ago

a major metroplitan newspaper selected a simple random sample of 1,600 readers from their list of 100,000 subscribers. they aske

Arisa [49]

Answer:

The 99% confidence interval for the proportion of readers who would like more coverage of local news is (0.3685, 0.4315).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.