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3 years ago

Hellllllpppppopp plzzzzz i dont know how to do this

Mathematics

1 answer:

Ronch [10]3 years ago

3 0

Answer:

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Step-by-step explanation:

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Helppppppppppppppppppp

ryzh [129]

Answer:

Step-by-step explanation:

You just substitute the values in for A and B and then you simplify

The 10 A would be 10x17 which is 170 and then plus B, which we know is 12, so it would equal 182

hope this helped

3 0

3 years ago

Read 2 more answers

Identify which formula we use to prove certain properties of polygons.

stepan [7]

Step-by-step explanation:

1. Both pairs of opposite sides are congruent Distance Formula

2. Both pairs of opposite sides are parallel Slope Formula

3.Diagonals are congruent Distance Formula

4. Diagonals bisect each other Midpoint Formula

5.Diagonals are perpendicular Slope Formula

6.Sides of polygon create a right angle Slope Formula

7.A triangle is isosceles / equilateral Distance Formula

6 0

3 years ago

HELPPPPPP!!!!!!!!!!!!!!!!!!!!!!!!!!!

dusya [7]

Answer:

sorry i dont understand

Step-by-step explanation:

7 0

3 years ago

Find the GCF of each pair of monominals 54gh,72g

wel

18 is the answer I believe hopefully this is correct

7 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