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3 years ago

How to set up a problem for law of sines when m

1 answer:

8 0

I think you meant to have more of a problem stated.
Basically, you use the Law of Sines when you have 2 angles but the length of only one side.

The formula for this law is
a / sine(A) = b / sine(B) = c / sine(C)

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Please answer quick!

inessss [21]

The answer should be the 3rd one

5 0

3 years ago

What is the area of the figure?

Kipish [7]

Answer:

17.98 square meters

Step-by-step explanation:

find the area of rectangle

length x width

6.2x1.8= 11.16

find area of triangle

base x height x 1/2

base= 6.2

height= 4-1.8= 2.2

area=2.2 x 6.2 x 1/2 = 6.82

add both together to get area of figure

6.82 + 11.16 = 17.98

5 0

3 years ago

What is the volume of this cube?

Annette [7]

Answer:

show us

Step-by-step explanation:

8 0

2 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

2 years ago

The number of students in a school increased by 25%between last year and this year. Last year there were 250 students in the sch

Ber [7]

I believe that is 275 because percent is just out of one hundred

8 0

3 years ago