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Makovka662 [10]
3 years ago
5

By what percent will the product of two numbers change, if the first number increases by 70%, while the second number decreases

by 40%.
Mathematics
1 answer:
FinnZ [79.3K]3 years ago
5 0

Answer:

2%

Step-by-step explanation:

Let x be the first number

It increases by 70 %

The new number is

m = x+ .70x

   = 1.7x

Let y be the second number

It decreases by 40 %

The new number is

n =y - .40 y

  = .6y


The product of the original numbers is xy

The product of the new number is

mn = (1.7x * .6y) = 1.02xy

The new number is larger than the old number so it is an increase.

Percent increase is  new - original divided by original  times 100%

Percent increase = (1.02 xy - xy)

                                -----------------  * 100%

                                 xy  

                           = .02 xy

                                ------- * 100 %

                                 xy  

                            = .02 * 100 %

                             = 2%



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Read 2 more answers
1 + tanx / 1 + cotx =2
Lera25 [3.4K]

Answer:

x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

Step-by-step explanation:

Solve for x:

1 + cot(x) + tan(x) = 2

Multiply both sides of 1 + cot(x) + tan(x) = 2 by tan(x):

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Subtract 2 tan(x) from both sides:

1 - tan(x) + tan^2(x) = 0

Subtract 1 from both sides:

tan^2(x) - tan(x) = -1

Add 1/4 to both sides:

1/4 - tan(x) + tan^2(x) = -3/4

Write the left hand side as a square:

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Take the square root of both sides:

tan(x) - 1/2 = (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

tan(x) = 1/2 + (i sqrt(3))/2 or tan(x) - 1/2 = -(i sqrt(3))/2

Take the inverse tangent of both sides:

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or tan(x) - 1/2 = -(i sqrt(3))/2

Add 1/2 to both sides:

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or tan(x) = 1/2 - (i sqrt(3))/2

Take the inverse tangent of both sides:

Answer:  x = tan^(-1)((i sqrt(3))/2 + 1/2) + π n_1 for n_1 element Z

or x = tan^(-1)(-(i sqrt(3))/2 + 1/2) + π n_2 for n_2 element Z

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