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3 years ago

12x-5y=20 y=x+4 3x+y=3 x=-y+3

1 answer:

4 0

Standard form (which I assume is what they want you to turn these into) is y=mx+b.

12x-5y=20 is y=2.4x-4.
Subtract 12x from both sides, -5y=-12x+20.
Multiply both sides by -1, 5y=12x-20.
Divide both sides by 5, y=2.4x-4.

y=x+4 is already in standard form- I'm not quite sure what to do.

3x+y=3 is y=-3x+3.
Subtract 3x from both sides, y=-3x+3.

x=-y+3 is y=-x+3
Ad y to both sides, y+x=3.
Subtract x from both sides, y=-x+3.

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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,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

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3 years ago

If the distance between A(0, 4) and B(3,a) is 5 units then find the value of a.

Anna007 [38]

Answer:

Step-by-step explanation:

a = 0

So that AOB makes a right triangle with sides 3-4-5

4 0

2 years ago

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The diameter of a sphere is 4 centimeters.

ELEN [110]

The volume of a sphere refers to the number of cubic units that will exactly fill a sphere. The volume of a sphere can be found or calculate by using the formula V=4/3πr^3, where r represents the radius of the figure.

In this exercise is given that a sphere has a radius of 4 centimeters and it is asked to find its volume and use 3.14 as the value of π or pi. The first step would be substitute the values into the previous mention formula.

V=4/3πr^3
V=4/3(3.14)(4 cm)^3
V=4/3)(3.14)(64 cm^3)
V=267.9 cm^3
The volume of the sphere is 267.9 cubic centimeters.

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3 years ago

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irina [24]

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MatroZZZ [7]

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