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Murljashka [212]

2 years ago

8

given triangle abc is similar to triangle efg, with medians cd and gh respectively, ac=9, eg=6, cd=x+2, gh=2x-4. find the length

of cd

1 answer:

Leno4ka [110]2 years ago

3 0

Answer:

X= 9//

Step-by-step explanation:

abc = efg

ac + cd = eg + gh

9 + (x+2) = 6 + (2x-4)

9 - 6 = 2x-4 - (x+2)

3 = x - 6

3+6 = x

9 = x//

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$(1\frac{1}{2},1)$ - False

$(4,6)$ - True

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$(0,0)$ -- True

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Step-by-step explanation:

The points are

$(1\frac{1}{2},1)$ , $(4,6)$ , $(18,12)$ , $(0,0)$ and $(2\frac{1}{2},3\frac{3}{4})$ ---- missing from the question

Given

$y = 1\frac{1}{2}x$

Required

Determine if each of the points would be on $y = 1\frac{1}{2}x$

To do this, we simply substitute the value of x and of each point in $y = 1\frac{1}{2}x$ .

(a) $(1\frac{1}{2},1)$

In this case;

$x = 1\frac{1}{2}$ and $y = 1$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 1\frac{1}{2}$

$y = \frac{3}{2} * \frac{3}{2}$

$y = \frac{9}{4}$

$y = 2\frac{1}{4}$

<em>The point </em> $(1\frac{1}{2},1)$ <em> won't be on the graph because the corresponding value of y for </em> $x = 1\frac{1}{2}$ <em> is </em> $y = 2\frac{1}{4}$ <em></em>

(b) $(4,6)$

In this case;

$x = 4$

$y = 6$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 4$

$y = \frac{3}{2} * 4$

$y = \frac{3* 4}{2}$

$y = \frac{12}{2}$

$y = 6$

<em>The point </em> $(4,6)$ <em> would be on the graph because the corresponding value of y for </em> $x = 4$ is $y = 6$

(c) $(18,12)$

In this case:

$x = 18;y = 12$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 18$

$y = \frac{3}{2} * 18$

$y = \frac{3* 18}{2}$

$y = \frac{54}{2}$

$y = 27$

<em>The point </em> $(18,12)$ <em> wouldn't be on the graph because the corresponding value of y for </em> $x = 18$ <em> is </em> $y = 12$ <em></em>

(d) $(0,0)$

In this case;

$x =0; y = 0$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 0$

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<em>The point </em> $(0,0)$ <em> would be on the graph because the corresponding value of y for </em> $x = 0$ is $y = 0$

(e) $(2\frac{1}{2},3\frac{3}{4})$

In this case:

$x = 2\frac{1}{2}$ ; $y = 3\frac{3}{4}$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 2\frac{1}{2}$

$y = \frac{3}{2} * \frac{5}{2}$

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$y = 3\frac{3}{4}$

<em>The point </em> $(2\frac{1}{2},3\frac{3}{4})$ <em> would be on the graph because the corresponding value of y for </em> $x = 2\frac{1}{2}$ is $y = 3\frac{3}{4}$

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