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3 years ago

How many times can 8.5 go into 147.90

Mathematics

1 answer:

Inga [223]3 years ago

3 0

Answer:

18.38

Step-by-step explanation:

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What is the period of the sinusoidal function? Enter your answer in the box.

muminat

Answer:

ANSWER:

The period of the sinusoidal function is 2π.

Step-by-step explanation:

EXPLANATION:

The sinusoidal function is a periodic function.It goes one unit up and one unit down with an amplitude of one and it repeats itself after this time interval.So,the period of the sine function is 2π

A sine or sinusoidal loop is a perpetual swing. It is called after the role sine. It happens frequently in tentative and practiced math, science, physics, engineering, signal processing, and other disciplines. Its utmost fundamental pattern as a role of time (t).

3 0

3 years ago

Triangle ABC is a right triangle what is the realationship between angles A and B

abruzzese [7]

Answer:

see explanation

Step-by-step explanation:

Complementary angles sum to 90°

∠A and ∠Bare complementary angles as they sum to 90°

59° + 31° = 90°

8 0

3 years ago

Please help! Graph the line for y+1=−3/5(x−4) on the coordinate plane. I just need help with the coordinates.

otez555 [7]

Answer:

SLOPE: - 3/5

Y-INTERCEPT: 7/5

Step-by-step explanation: Check picture :)

5 0

3 years ago

Ok real quick question how does your teacher give you a D on one of your assignments after they just signed you up for an honor

grandymaker [24]

Answer:

when the signed you up for the honor class they probably expected you to know more

4 0

3 years ago

Read 2 more answers

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