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My name is Ann [436]
2 years ago
13

Type in uwu text. do it.

Mathematics
2 answers:
olganol [36]2 years ago
5 0

Answer:

uwu hwai i wove ur pfp :)

Explain:

N e e d e d 20 c h a r a c t e r s

Afina-wow [57]2 years ago
4 0

Answer:

UwU

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
Evaluate the limit and express your answer in simplest form<br> tar<br> 5/4<br> 4/5<br> 2/3<br> 2/5
muminat

Answer: how do we have the same question need the answer too

Step-by-step explanation:

5 0
3 years ago
PLEASE HELP ASAP
Tpy6a [65]
The area of the polygon with vertices W (1, 1), X (4, 4), Y (7, 1), and S (4, −8) is 36.
Have a great day! :D
6 0
2 years ago
How to factor 6n2 - 6n - 12
alexandr402 [8]

We see that in these 3 terms, 6 is a common factor. So, let's factor out a 6:

6n^2-6n-12 = 6(n^2-n-2)

We can set this equal to 0 and factor by using the quadratic formula which is:

x = \frac{-b \pm \sqrt{b^2-4ac} }{2a}

So, let's do just that:

6(n^2-n-2) = 0

Note that the 6 goes away if you divide both sides by 6. In this case, a = 1, b = -1, and c = -2. Let's plug that into the quadratic equation:

\frac{1 \pm \sqrt{(-1)^2-4(1)(-2)} }{2(1)} = \frac{-1 \pm \sqrt{1-(-8)} }{2}

\frac{-1 \pm \sqrt{9} }{2} = {1, -2}

So, we can write this as:

6n^2 - 6n - 12 = 6(n-1)(n+2)

Notice that the 6 comes back because it was only temporarily mad. And that the roots have opposite signs in the parentheses because to find the roots, you need to set each parentheses equal to 0 and solve for n. With that in mind, your final answer is 6(n-1)(n+2). Hope I could help you!

7 0
3 years ago
Solve the inequality. Graph the solution set.<br><br> 2r−9≤−6
Ahat [919]
We have that

2r−9≤−6------> 2r ≤ -6+9-------> 2r ≤  3-----> r ≤ 1.5

the solution is the interval  (-∞, 1.5]

using a graph tool
see the attached figure

5 0
3 years ago
Read 2 more answers
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