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aleksandr82 [10.1K]
2 years ago
10

If DF=7x-3 and EG=5x+9, find the value of X

Mathematics
1 answer:
PilotLPTM [1.2K]2 years ago
3 0

Answer:

x=6

Step-by-step explanation:

Using the given measurements of the angles and the lengths of the sides, we can determine that this figure must be a <u>rectangle</u>.

Diagonals of a rectangle are <em>always</em> equivalent/congruent.

We can use this information to set up an equation:

DF=EG\\7x-3=5x+9

Add 3 to both sides:

7x-3+3=5x+9+3\\7x=5x+12

Subtract 5x from both sides:

7x-5x=5x-5x+12\\2x=12

Divide both sides by 2

\frac{2x}{2}=\frac{12}{2}\\x=6

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Answer:

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Step-by-step explanation:

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2 years ago
Find a positive number for which the sum of it and its reciprocal is the smallest​ (least) possible. Let x be the number and let
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Answer:

S(x) = x + \frac{1}{x} --- Objective function

Interval = \{x:x=1\}

Step-by-step explanation:

Given

Represent the number with x

The required sum can be represented as:

x + \frac{1}{x}

Hence, the objective function is:

S(x) = x + \frac{1}{x}

To get the the interval, we start by differentiating w.r.t x

<em>Using first principle, this gives:</em>

S'(x) = 1 - \frac{1}{x^2}

Equate S'(x) to 0 in order to solve for x

0 = 1 - \frac{1}{x^2}

Subtract 1 from both sides

0 -1 = 1 -1 - \frac{1}{x^2}

-1 = - \frac{1}{x^2}

Multiply both sides by -1

1 = \frac{1}{x^2}

Cross Multiply

x^2 * 1 = 1

x^2  = 1

Take positive square root of both sides because x is positive

\sqrt{x^2} = \sqrt{1

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Representing x using interval notation, we have

Interval = \{x:x=1\}

To get the smallest sum, we substitute 1 for x in S(x) = x + \frac{1}{x}

S(1) = 1 + \frac{1}{1}

S(1) = 1 + 1

S(1) = 2

<em>Hence, the smallest sum is 2</em>

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