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BaLLatris [955]
3 years ago
15

If angle 1 and angle 7 are vertical angles and angle 1 and angle 2 are supplementary adjacent and the measure of angle 2 is 123

degrees. what is the measure of angle 7?
Mathematics
1 answer:
JulsSmile [24]3 years ago
7 0
Answer:  m∡7 = 57° .
_________________________________

Explanation:
____________________
 m∡7 = m∡1 ;  SInce  these angles are vertical; and vertical angles are congruent.

m∡1 + m∡2 = 180 ; since these angles are supplementary; and by                                            definition; supplementary angles "add up" to 180.

Given: m∡2 = 123 ;  

m∡1 + 123 = 180 ;

Since m∡1 = m∡7 ; and we wish to find: "m∡7" ;  we can rewrite as:
_______________________________
m∡7 + 123 = 180 ; solve for "m∡7" ;
_______________________________
Subtract "123" from EACH SIDE of the equation:
_________________________________________
  m∡7 + 123 - 123 = 180 - 123 ;
_________________________________________
  m∡7 = 57° .
_________________________________________
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3 0
3 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

x = 1

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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