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mariarad [96]
3 years ago
7

Find the sum. 5 2a a? + 2a + 1 + a2 + 4a + 3

Mathematics
2 answers:
Vinvika [58]3 years ago
7 0

Answer:

=  \frac{2 {a}^{2}  + 11a + 5}{  {(a + 1)}^{2}(a + 3)} \\

Step-by-step explanation:

\frac{2a}{ {a}^{2}  + 2a + 1} +  \frac{5}{ {a}^{2} + 4a + 3 } \\  \frac{2a}{(a + 1)(a + 1)}    +  \frac{5}{(a + 1)(a + 3)}  \\   \frac{2a}{ {(a + 1)}^{2} }  +  \frac{5}{(a + 1)(a + 3)}  \\  \frac{2a(a + 3) + 5(a + 1)}{ {(a + 1)}^{2}(a + 3) }  \\  \frac{2 {a}^{2} + 6a + 5a + 5 }{ {(a + 1)}^{2}(a + 3)} \\  =  \frac{2 {a}^{2}  + 11a + 5}{  {(a + 1)}^{2}(a + 3)}

Salsk061 [2.6K]3 years ago
5 0

\frac{2a}{a^2+2a+1}+\frac{5}{a^2+4a+3}

Factor the denominators.

\frac{2a}{\left(a+1\right)^2}+\frac{5}{\left(a+1\right)\left(a+3\right)}

Adjust fractions based on LCM.

\frac{2a\left(a+3\right)}{\left(a+1\right)^2\left(a+3\right)}+\frac{5\left(a+1\right)}{\left(a+1\right)^2\left(a+3\right)}

Denominators are same, so add the fractions.

\frac{2a\left(a+3\right)+5\left(a+1\right)}{\left(a+1\right)^2\left(a+3\right)}

Expand the numerator.

\frac{2a^2+11a+5}{\left(a+1\right)^2\left(a+3\right)}

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If triangle ABC is reflected over the y‐axis, reflected over the x‐axis, and rotated 180 degrees, where will point A' lie?
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Given that ABC is a triangle and is reflected over the y - axis, reflected over the x - axis and rotated 180°

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<u>Reflection over the y - axis:</u>

The coordinates of the point A is (-2,1)

The transformation rule to reflect across the y - axis is (x,y)\rightarrow (-x,y)

Substituting the point (-2,1) in the rule, we get;

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<u>Reflection over the x - axis:</u>

The transformation rule to reflect across the x - axis is (x,y)\rightarrow (x,-y)

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