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3 years ago

Uestion Hel

Mathematics

1 answer:

Evgesh-ka [11]3 years ago

6 0

40ft hope this hellped ask again

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Find the equation of the axis of symmetry and the coordinates of the vertex of the graph of the function y = 4x2 + 5x – 1.

Tanya [424]

Answer:

The option u chose is the correct one

4 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

3 years ago

I need help getting the answer

djverab [1.8K]

1.) $0.22
2.) $0.19
you just divide the price by the # of tortillas

4 0

3 years ago

Cailyn got 2 out of the 3 questions <br> correct on her test. What is her grade?

Stolb23 [73]

Answer: 80

Step-by-step explanation:

# Wrong Grade

1 90

2 80

3 70

4 60

3 0

2 years ago

For the following figure, complete the statement for the specified points. Points R, T, B, and C are _____. collinear coplanar b

melamori03 [73]

Answer:

Coplanar

Step-by-step explanation:

We know that the collinear points are the points which together make a straight line. From the diagram, it is clearly visible that points R, T, and B make a straight line but C doesn't follow the path of RTB.

So it not collinear.

We know that that coplanar points are the points which together make a closed shapes in two-dimensions. From the diagram, it is clearly visible that points R, T, B, and C make a closed shape like a triangle.

So it is coplanar.

6 0

4 years ago