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Mathematics

1 answer:

fomenos3 years ago

6 0

Answer:

Since 1 in : 4 yd i just multiply 12 by 4 so I get 48 so 48 square yards is your answer

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Find the product. (r s)2 r² 2rs s r² 2rs s² r² 2rs 2s².

sveta [45]

The product of the given expression is $\rm 2r^3+ s^3r^3+ 2r^3s + sr^3+ 2s^3+ 6rs$ .

Given

The product of $\rm 2r^2(r + s) + 2rs + s^2 + 2rs + s^2r^2+ 2rs + 2s^2$ .

<h3>What is multiplication?</h3>

Multiplication is the process of calculating the product of two or more numbers.

The multiplication of numbers say, ‘a’ and ‘b’, is stated as ‘a’ multiplied by ‘b’.

The product of $\rm 2r^2(r + s) + 2rs + s^2 + 2rs + s^2r^2+ 2rs + 2s^2$ is given by;

$\rm 2r^2(r + s) + 2rs + s^2 + 2rs + s^2r^2+ 2rs + 2s^2\\\\(2r^3 + 2r^2s) + 2rs + 2rs + 2rs + s^2r^2 + sr^2 + 2s^2\\\\2r^3+ 2r^2s + 6rs + s^2r^2 + sr^2+ 2s^2$

$\rm 2r^3+ s^3r^3+ 2r^3s + sr^3+ 2s^3+ 6rs$

Hence, the product of the given expression is $\rm 2r^3+ s^3r^3+ 2r^3s + sr^3+ 2s^3+ 6rs$ .

To know more about Multiplication click the link given below.

brainly.com/question/19943359

6 0

2 years ago

How do I solve this?

Verdich [7]

Consider the point P where the gears meet. When the smaller gear rotates clockwise, the larger one will rotate counterclockwise.

Through one rotation of the smaller gear, P will have traveled the circumference of the smaller gear, which is $8\pi$ in.

At the same time, a point P' on the larger gear traverses the same distance along the larger gear's circumference. This point traces out an arc that is subtended by some angle $\theta$ . The arc is as long as the smaller gear's circumference.

The measure of a circle's interior angle subtended by an arc is proportional to a complete revolution, i.e. an angular displacement of $2\pi$ radians:

$\dfrac{14\pi\,\mathrm{in}}{2\pi\,\mathrm{rad}}=\dfrac{8\pi\,\mathrm{in}}\theta\implies\theta=\dfrac{8\pi}7\,\mathrm{rad}\approx205.7^\circ$

For part 2, we apply the same reasoning to the larger gear. In one full rotation of the larger gear, the point P' traverses the circumference $14\pi$ in, and so does the point P on the smaller gear.