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IRINA_888 [86]
1 year ago
10

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

Mathematics
1 answer:
sveta [45]1 year ago
6 0

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

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Step-by-step explanation:

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4 0
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In 3-4 sentences, describe how you would find a line parallel to the line 2x + 5y = 15 that goes through the point (-10, 1). Be
larisa86 [58]

the parallel line is 2x+5y+15=0.

Step-by-step explanation:

ok I hope it will work

soo,

Solution

given,

given parallel line 2x+5y=15

which goes through the point (-10,1)

now,

let 2x+5y=15 be equation no.1

then the line which is parallel to the equation 1st

2 x+5y+k = 0 let it be equation no.2

now the equation no.2 passes through the point (-10,1)

or, 2x+5y+k =0

or, 2*-10+5*1+k= 0

or, -20+5+k= 0

or, -15+k= 0

or, k= 15

putting the value of k in equation no.2 we get,

or, 2x+5y+k=0

or, 2x+5y+15=0

which is a required line.

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10 is double 5 so,
32.5 multiplied by 2 is: 65
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