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3 years ago

I WILL REWARD BRAINLIEST

Mathematics

2 answers:

Alex Ar [27]3 years ago

8 0

It doesnt equal either of them 10 only = 10

svet-max [94.6K]3 years ago

3 0

3,628,800
Ultimately you do this: 10*9*8*7*6*5*4*3*2*1

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Sarah's holiday wreath has 5 more than twice as many blue lights as green lights and 7 less than 5 times as many red lights as g

gregori [183]

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4 0

3 years ago

You read 165 pages in 3 hours find the unit rate

Aliun [14]

Answer:

55 pages per hour

Step-by-step explanation:

You divide 165 by 3 = 55

8 0

3 years ago

A square sheet of paper measures 42 centimeters on each side. What is the length of the diagonal of this paper?

Illusion [34]

59.4cm
Use Pythagorean theorem to find the diagonal (^ means exponent)
a^2 + b^2 = c^2
42^2 + 42^2= c^2
1764+1764=c^2
3528=c^2
square root both sides
c= 59.39696962
round and you get 59.4 cm

5 0

4 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

What is the rule for the reflection?

Dvinal [7]

Answer:

Incident ray, the reflection ray and the normal to thereflection surface at the point of the incidence lie in the same plane

5 0

3 years ago