By what percent will the product of two numbers decrease if one is decreased by 75% and the other is decreased by 20%?

The product of two numbers will decrease by 70%

Let the numbers be x and y.

So, the product of x and y is represented as: xy

If x is decreased by 75%, then it becomes x(1 - 75%)

If y is increased by 20%, then it becomes y(1 + 20%)

So, the product of the new numbers is:

Express percentage as decimals

This gives

Rewrite the above equation as:

Evaluate the products

Rewrite as:

Express 0.3 as 1- 0.7

Express 0.7 as percentage

Hence, the product of two numbers will decrease by 70%

Read more about percentage change at:

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80%

Step-by-step explanation:

Let the first number be "100" and the other number be "100" as well.

Their normal product would be:

100 * 100 = 10,000

Now, we need to decrease one by 75% and another by 20%. First, let's convert the percentages to decimals. Shown below:

75% = 75/100 = 0.75

20% = 20/100 = 0.2

First number is "100", so decrease it by 75% to get:

100 - 0.75(100)

= 100 - 75

= 25

Next number is also "100", so decrease it by 20% to get:

100 - 0.2(100)

= 100 - 20

= 80

The new product would eb:

80 * 25 = 2000

Original Product = 10,000

New Product = 2000

To find percentage decrease, we use the formula:

Lets substitute and find the

So, the new product is 80% decreased

mxb67

step-by-step explanation:

x^2 + 81

we can factor this using difference of square binomials.

rewrite the equation:

x^2 - (-81)

factor:

(a^2 - b^2) = (√a^2 + √b^2)(√a^2 - √b^2)

(x^2 - (-81) = (x + √-81)(x - √-81)

we can remove the - from under the radical by multiplying it by i.

(x^2 - (-81) = (x + i√81)(x - i√81)

(x^2 - (-81) = (x + 9i)(x - 9i)

we can check our answer by using foil:

x^2 - 9ix + 9ix - 81i^2

x^2 - 81^2

i^2 = -1

x^2 + 81