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zhenek [66]
1 year ago
6

Which value of x is in the solution set of 2x-3>11-5x

Mathematics
3 answers:
Serhud [2]1 year ago
8 0

7x>14,x=2። is right answer i think

kipiarov [429]1 year ago
4 0

Answer:

Value of x > 2 is in the solution set of 2x-3>11-5x which means the value of x can be any value greater than 2.

Step-by-step explanation:

We need to find the value of x for the equation 2x-3>11-5x

2x-3>11-5x

Adding +5x on both sides:

2x+5x-3>11-5x+5x

7x-3>11

Adding +3 on both sides

7x-3+3>11+3

7x>14

Dividing by 7 on both sides:

7x/7>14/7

x>2

Value of x > 2 is in the solution set of 2x-3>11-5x which means the value of x can be any value greater than 2.

Ierofanga [76]1 year ago
3 0

7x>14,x=2። is right answer i think

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Answer:

3 - (n - 1) = 1/2(3n - 4)

Step-by-step explanation:

We want to write three minus the difference of a number and one equals one-half of the difference of three times the same number and four as an equation.

Let the number be n.

The first part is: three minus the difference of a number and one:

3 - (n - 1)

The second part is: one-half of the difference of three times the same number and four:

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Now, let us equate the first and second parts:

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1 year ago
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A surveyor, Toby, measures the distance between two landmarks and the point where he stands. He also measured the angles between
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The Set Up:

x² = (Side1)² + (Side2)² - 2[(Side1)(Side2)]

Solution:

cos(Toby's Angle) • x² = 55² + 65² - 2[(55)(65)] cos(110°)

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The distance, x, between two landmarks is 69.31m.

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2 years ago
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Given:

Area of rectangle = 6n^4+20n^3+14n^2

Width of the rectangle is equal to the greatest common monomial factor of 6n^4, 20n^3,14n^2.

To find:

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Solution:

Width of the rectangle is equal to the greatest common monomial factor of 6n^4, 20n^3,14n^2 is

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20n^3=2\times 2\times 5\times n\times n\times n

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We know that, area of a rectangle is the product of its length and width.

Since, width of the rectangle is 2n^2, therefore length of the rectangle is (3n^2+10n+7).

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