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2 years ago

If line q bisects EG at point T, and if ET= 1/3 x and TG= x-2, then find EG

Mathematics

2 answers:

kvasek [131]2 years ago

8 0

Answer:

EG = 2 units

Step-by-step explanation:

Given that line q bisects EG at T , then

ET = TG ( substitute values )

$\frac{1}{3}$ x = x - 2 ( multiply through by 3 to clear the fraction )

x = 3x - 6 ( subtract x from both sides )

0 = 2x - 6 ( add 6 to both sides )

6 = 2x ( divide both sides by 2 )

3 = x

Then

ET = $\frac{1}{3}$ x = $\frac{1}{3}$ × 3 = 1

TG = x - 2 = 3 - 2 = 1

Thus

EG = ET + TG = 1 + 1 = 2 units

Rashid [163]2 years ago

5 0

EG = $\frac{4x}{3} - 2$

Step-by-step explanation:

( <em>First check the attached image for diagram</em> )

$EG=ET+TG$

Substituting the given values,

$EG = \frac{1}{3}x + (x - 2)$

Opening the brackets and multiplying,

$EG = \frac{1 \times x}{3} + x - 2$

$= > EG = \frac{x}{3} + x - 2$

Finding LCM,

$= > EG = ( \frac{x}{3} + \frac{x \times 3}{1 \times 3} ) - 2$

$= > EG = ( \frac{x}{3} + \frac{3x}{3}) - 2$

Combining them,

$EG =( \frac{x + 3x}{3}) - 2$

$EG = \frac{4x}{3} - 2$

Note:- T is midpoint of EG

because ET = TG

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patriot [66]

Answer:

Choice b.

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Step-by-step explanation:

The highest power of the variable $x$ in this polynomial is $2$ . In other words, this polynomial is quadratic.

It is thus possible to apply the quadratic formula to find the "roots" of this polynomial. (A root of a polynomial is a value of the variable that would set the polynomial to $0$ .)

After finding these roots, it would be possible to factorize this polynomial using the Factor Theorem.

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Similarly:

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By the Factor Theorem, if $x = x_{0}$ is a root of a polynomial, then $(x - x_0)$ would be a factor of that polynomial. Note the minus sign between $x$ and $x_{0}$ .

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2 years ago

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Kobotan [32]

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3 years ago

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andrey2020 [161]

Answer:

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