among all pairs of numbers whose difference is 10
,
find a pair whose product is as small as possible. what is the minimum product?

We want to find two numbers that have a difference of 10, such that its product is minimized.

We will find that the two numbers are 5 and -5, and the minimum possible product is p = -25.

Let's see how to solve this.

So we can define two numbers x and y, such that:

x - y = 10

Now we can write the product of these two numbers as:

p = x*y

From the first equation, we can isolate one of the two variables and write:

x = 10 + y

Now we can replace that in the product equation to get:

p = x*y = (10 + y)*y

p = 10*y + y^2

Now we want to minimize this quadratic equation. Notice that it has a positive leading coefficient, thus, the minimum will be at the vertex of the equation.

Remember that for the general quadratic equation:

Y = a*X^2 + b*X + c

The vertex is at:

X = -b/2a

Then in our case:

p = 10*y + y^2

The vertex is at:

y = -10/2*1 = -5

Then we got the value of y, now we can find the value of x:

x = 10 + y = 10 + (-5) = 5

x = 5

Then the pair of numbers is 5 and -5, and the minimum product is:

p = 5*(-5) = -25

If you want to learn more, you can read:

link

what does muffins have to do with anything?

step-by-step explanation:

use the power rule  

( a b ) n = a n b n

to distribute the exponent.

apply the product rule to  

3 x 2 y 3 . ( 3 x 2 ) 2 ( y 3 ) 2

apply the product rule to  

3 x 2 . 3 2 ( x 2 ) 2 ( y 3 ) 2

raise  

3

to the power of  

2 . 9 ( x 2 ) 2 ( y 3 ) 2 multiply the exponents in   ( x 2 ) 2

.

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 2 ⋅ 2 ( y 3 ) 2

multiply  

2   by   2 . 9 x 4 ( y 3 ) 2

multiply the exponents in  

( y 3 ) 2 .

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 4 y 3

⋅

2

multiply  

3   by   2 . 9 x 4 y 6

step-by-step explanation: