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rjkz [21]
3 years ago
13

Please help it is midpoint

Mathematics
2 answers:
Maslowich3 years ago
5 0

Answer:

mid \: point \: for \: x = \frac{x1 + x2}{2}  \\  =  \frac{4.6 + 12}{2}  \\ x =  \frac{16.6}{2}  \\ x = 8.3 \\ mid \: point \: of \: y =  \frac{y1 + y2}{2}  \\ y =  \frac{2.3 + 4}{2}  \\  \frac{6.3}{2}  = 3.15 \\ then(x \: y) = (8.3 \: 3.15) \\ thank \: you

inna [77]3 years ago
3 0

Answer:

(8.3, 3.15 )

Step-by-step explanation:

Given endpoints (x₁, y₁ ) and (x₂, y₂ ) then the midpoint is

( \frac{x_{1}+x_{2}  }{2} , \frac{y_{1}+y_{2}  }{2} )

Here (x₁, y₁ ) = (4.6, 2.3) and (x₂, y₂ ) = (12, 4) , then

midpoint = ( \frac{4.6+12}{2} , \frac{2.3+4}{2} ) = ( \frac{16.6}{2} , \frac{6.3}{2} ) = ( 8.3, 3.15 )

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Answer:

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Step-by-step explanation:

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4 years ago
What is the exact value of the expression square foot of 486 − square root of 24+ square root of 6 ?
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3 years ago
Subtract. <br> –84 – 55<br> a. –29<br> b. –139<br> c. 29<br> d. 139
Hitman42 [59]
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8 0
3 years ago
Simplify (x + 2/ x^2 + 2x -3) / (x + 2/x^2 - x)
rjkz [21]

Answer:

The simplest form is x/(x + 3)

Step-by-step explanation:

* To simplify the rational Expression lets revise the factorization

  of the quadratic expression

*  To factor a quadratic in the form x² ± bx ± c:

- First look at the c term  

# If the c term is a positive number, and its factors are r and s they

  will have the same sign and their sum is b.

#  If the c term is a negative number, then either r or s will be negative

   but not both and their difference is b.

- Second look at the b term.  

# If the c term is positive and the b term is positive, then both r and

  s are positive.  

Ex: x² + 5x + 6 = (x + 3)(x + 2)  

# If the c term is positive and the b term is negative, then both r and s

  are negative.  

Ex:  x² - 5x + 6 = (x -3)(x - 2)

# If the c term is negative and the b term is positive, then the factor

  that is positive will have the greater absolute value. That is, if

  |r| > |s|, then r is positive and s is negative.  

Ex: x² + 5x - 6 = (x + 6)(x - 1)

# If the c term is negative and the b term is negative, then the factor

  that is negative will have the greater absolute value. That is, if

  |r| > |s|, then r is negative and s is positive.

Ex: x² - 5x - 6 = (x - 6)(x + 1)

* Now lets solve the problem

- We have two fractions over each other

- Lets simplify the numerator

∵ The numerator is \frac{x+2}{x^{2}+2x-3}

- Factorize its denominator

∵  The denominator = x² + 2x - 3

- The last term is negative then the two brackets have different signs

∵ 3 = 3 × 1

∵ 3 - 1 = 2

∵ The middle term is +ve

∴ -3 = 3 × -1 ⇒ the greatest is +ve

∴ x² + 2x - 3 = (x + 3)(x - 1)

∴ The numerator = \frac{(x+2)}{(x+3)(x-2)}

- Lets simplify the denominator

∵ The denominator is \frac{x+2}{x^{2}-x}

- Factorize its denominator

∵  The denominator = x² - 2x

- Take x as a common factor and divide each term by x

∵ x² ÷ x = x

∵ -x ÷ x = -1

∴ x² - 2x = x(x - 1)

∴ The denominator = \frac{(x+2)}{x(x-1)}

* Now lets write the fraction as a division

∴ The fraction = \frac{x+2}{(x+3)(x-1)} ÷ \frac{x+2}{x(x-1)}

- Change the sign of division and reverse the fraction after it

∴ The fraction = \frac{(x+2)}{(x+3)(x-1)}*\frac{x(x-1)}{(x+2)}

* Now we can cancel the bracket (x + 2) up with same bracket down

 and cancel bracket (x - 1) up with same bracket down

∴ The simplest form = \frac{x}{x+3}

5 0
4 years ago
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