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Fittoniya [83]
2 years ago
15

Solve 3x^2+19x +6 please help

Mathematics
1 answer:
slega [8]2 years ago
8 0

Answer:

(3x + 1)(x + 6)

3x^2 + 19x + 6

3x^2 + 18x + x + 6

3x^2 + 18x + x + 6

3x(x + 6) + x + 6

3x(x + 6) + x + 6

(3x + 1)(x + 6)

Haven't done this in a pretty long time so I might be wrong

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Step-by-step explanation:

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parallelogram PQRS has PQ=RS=8 cm and diagonal QS= 10 cm. point F is on RS exactly 5 cm from S. let T be the intersection of PF
Doss [256]

<u>Solution-</u>

Given that,

In the parallelogram PQRS has PQ=RS=8 cm and diagonal QS= 10 cm.

Then considering ΔPQT and ΔSTF,

1-    ∠FTS ≅ ∠PTQ            ( ∵ These two are vertical angles)

2-   ∠TFS ≅ ∠TPQ            ( ∵ These two are alternate interior angles)

3-   ∠TSF ≅ ∠TQP            ( ∵ These two are also alternate interior angles)

<em>If the corresponding angles of two triangles are congruent, then they are said to be similar and the corresponding sides are in proportion.</em>

∴ ΔFTS ∼ ΔPTQ, so corresponding side lengths are in proportion.

\Rightarrow \frac{PQ}{FS} =\frac{TQ}{TS} =\frac{TP}{TF}

As QS = TQ + TS = 10 (given)

If TS is x, then TQ will be 10-x. Then putting these values in the equation

\Rightarrow \frac{PQ}{FS} =\frac{TQ}{TS}

\Rightarrow \frac{8}{5} =\frac{10-x}{x}

\Rightarrow x=3.85

∴ So TS = 3.85 cm and TQ is 10-3.85 = 6.15 cm




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