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Effectus [21]
3 years ago
6

I need help its for Advanced Algebra

Mathematics
1 answer:
SCORPION-xisa [38]3 years ago
4 0
Just here for points sorry
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s=25

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If two angles are complements of each other then each angle is
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each angle is less than 90 degrees. angle 1+angle2=90 degrees

Step-by-step explanation:

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❊ Simplify :
Nataliya [291]
<h3>Need to Do :- </h3>
  • To simplify the given expression .

\red{\frak{Given}}\Bigg\{ \sf \dfrac{x - 1}{ {x}^{2} - 3x + 2} + \dfrac{x - 2}{ {x}^{2} - 5x + 6 } + \dfrac{x - 5}{ {x}^{2} - 8x + 15 }

\rule{200}4

\sf\longrightarrow \small  \dfrac{x - 1}{ {x}^{2} - 3x + 2} + \dfrac{x - 2}{ {x}^{2} - 5x + 6 } + \dfrac{x - 5}{ {x}^{2} - 8x + 15 } \\\\\\\sf\longrightarrow \small  \dfrac{ x-1}{x^2-x -2x +2} +\dfrac{ x-2}{x^2-3x-2x+6} +\dfrac{ x -5}{x^2-5x -3x + 15 } \\\\\\\sf\longrightarrow\small \dfrac{ x -1}{ x ( x - 1) -2(x-1) } +\dfrac{ x-2}{x ( x -3) -2( x -3)} +\dfrac{ x -5}{ x(x-5) -3( x -5) }  \\\\\\\sf\longrightarrow \small \dfrac{ x -1}{ ( x-2) (x-1) } +\dfrac{ x-2}{( x -2)(x-3) } +\dfrac{ x -5}{ (x-3)(x-5)  } \\\\\\\sf\longrightarrow\small \dfrac{ 1}{ x-2} +\dfrac{ 1}{ x -3} +\dfrac{1}{ x -3}   \\\\\\\sf\longrightarrow   \small  \dfrac{1}{x-2} +\dfrac{2}{x-3}  \\\\\\\sf\longrightarrow   \small \dfrac{ x-3 +2(x-2)}{ ( x -3)(x-2) }  \\\\\\\sf\longrightarrow   \small \dfrac{ x - 3 +2x -4 }{ (x-3)(x-2) }     \\\\\\\sf\longrightarrow   \underset{\blue{\sf Required \ Answer  }}{\underbrace{\boxed{\pink{\frak{  \dfrac{ 3x -7}{ ( x -2)(x-3) } }}}}}

\rule{200}4

5 0
2 years ago
Read 2 more answers
Please explain how they plugged the removable discontinuity in this question.
kondor19780726 [428]

In the limit

\displaystyle \lim_{x\to c} f(x)

we're interested in the value that f(x) converges to as x gets closer to c. So in fact x\neq c.

In the given example, f(x) is factorized to reveal a common factor of x-1 in the numerator and denominator. We have x\neq1 if x\to1, so x-1\neq0 so we can simplify

\dfrac{x-1}{x-1} = 1

and *remove* the discontinuity.

Then

\displaystyle \lim_{x\to1} \frac{2(x-1)}{(x+1)(x-1)} = \lim_{x\to1} \frac2{x+1} = \frac2{1+1} = 1

8 0
1 year ago
Trina, James, and Michael are training for a 10-kilometer race. After 30 minutes of a training run, Michael lead the group, and
Llana [10]

Answer:

1/3

Step-by-step explanation:

1/2 James' fraction of Michael

2/3 Trina's fraction of James

1/2 x 2/3 = 1/3

5 0
3 years ago
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