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3 years ago

1/3 = /6 - 1/6 = /6 How do you find the unlike denominator. Can you explain

Mathematics

1 answer:

lakkis [162]3 years ago

5 0

Answer:

Okay So I see you need to change 1/3 to a denominator of 6

You don't do anything to the -1/6

But to change the 1/3 you need to multiply 3 and 1 by two, which is 2/6

And the -1/6 stays the same

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What is UT?

kenny6666 [7]

In the right triangle given, the length of UT using the Pythagorean theorem is: UT = 15.

<h3>What is the Pythagorean Theorem?</h3>

Pythagorean theorem is given as: c² = a² + b², where a and b are the legs of a right triangle, and c is the hypotenuse (the side opposite the right angle).

Given:

c = TV = 17

a = UT = ?

b = UV = 8

Plug in the values

17² = UT² + 8²

UT = √(17² - 8²)

UT = 15.

Thus, in the right triangle given, the length of UT using the Pythagorean theorem is: UT = 15.

Learn more about the Pythagorean theorem on:

brainly.com/question/21332040

8 0

2 years ago

Answer this question to get marked as brainliest!!!!!

timurjin [86]

(15^3)^3

the rest are equal to 15^6:)

4 0

2 years ago

Please show me how to do #11.

Elodia [21]

Do number 11 in your own way.

5 0

3 years ago

What is the approximate volume of the cylinder? Use 3.14 for п.

Veseljchak [2.6K]

Given:

Given that the radius of the cylinder is 4 cm.

The height of the cylinder is 9 cm.

We need to determine the volume of the cylinder.

Volume of the cylinder:

The volume of the cylinder can be determined using the formula,

$V=\pi r^2 h$

where r is the radius and h is the height of the cylinder.

Substituting π = 3.14, r = 4 and h = 9 in the above formula, we get;

$V=(3.14)(4)^2(9)$

$V=(3.14)(16)(9)$

$V=452.16 \ cm^3$

Thus, the volume of the cylinder is 452.16 cm³

Hence, Option B is the correct answer.

4 0

3 years ago

A) Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given cur

Leno4ka [110]

(a) See the attached sketch. Each shell will have a radius y chosen from the interval [2, 4], a height of x = 2/y, and thickness ∆y. For infinitely many shells, we have ∆y converging to 0, and each super-thin shell contributes an infinitesimal volume of

2π (radius)² (height) = 4πy

Then the volume of the solid is obtained by integrating over [2, 4]:

$\displaystyle 4\pi \int_2^4 y\,\mathrm dy = 2\pi y^2\bigg|_{y=2}^{y=4} = 2\pi (4^2-2^2) = \boxed{24\pi}$

(b) See the other attached sketch. (The text is a bit cluttered, but hopefully you'll understand what is drawn.) Each shell has a radius 9 - x (this is the distance between a given x value in the orange shaded region to the axis of revolution) and a height of 8 - x ³ (and this is the distance between the line y = 8 and the curve y = x ³). Then each shell has a volume of

2π (9 - x)² (8 - x ³) = 2π (648 - 144x + 8x ² - 81x ³ + 18x ⁴ - x ⁵)

so that the overall volume of the solid would be

$\displaystyle 2\pi \int_0^2 (648-144x+8x^2-81x^3+18x^4-x^5)\,\mathrm dx = \boxed{\frac{24296\pi}{15}}$

I leave the details of integrating to you.

3 0

2 years ago