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Sonbull [250]
3 years ago
8

Point d is the centroid of triangle abc use the given information to find x. GD=2x-8 and GC=3x+3

Mathematics
2 answers:
BigorU [14]3 years ago
5 0
The centroid D cuts every median in the ratio 2:1. The distance between a vertex and the centroid is twice the distance between the centroid and the midpoint of the opposite side.

Therefore: |DC| = 2|GD|
|GC| = |GD| + |DC| = |GD| + 2|GD| = 3|GD|

so... we have:

3x + 3 = 3(2x-8)
3x + 3 = 3 × 2x - 3 × 8
3x + 3 = 6x - 24    |<em>subtract 3 from both sides</em>
3x = 6x -27      |<em>subtract 6x from both sides</em>
-3x = -27    |<em>divide both sides by (-3)</em>

x = 9


sertanlavr [38]3 years ago
4 0
Answer and workings in the attachments below.

x=9.

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Step-by-step explanation:

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pickupchik [31]

1.125

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5. One kilogram is 2.2 pounds. Complete the tables. What is the interpretation of the constant of proportionality in each case?​
Sergeeva-Olga [200]

Answer:

There is 0.5 kiligrams in one pound

Step-by-step explanation:

4.98952

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6 0
2 years ago
Polygon q is a scaled copy of polygon p using a scale factor of 1/2 polygon q’s area is what fraction of polygon p’s area?
san4es73 [151]

Answer:

Polygon q’s area is one fourth of polygon p’s area

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its areas is equal to the scale factor squared

Let

z-----> the scale factor

x-----> polygon q’s area

y-----> polygon p’s area

so

z^{2} =\frac{x}{y}

In this problem we have

z=\frac{1}{2}

substitute

(\frac{1}{2})^{2} =\frac{x}{y}

(\frac{1}{4}) =\frac{x}{y}

x=\frac{1}{4}y

therefore

Polygon q’s area is one fourth of polygon p’s area

3 0
3 years ago
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Can the mean value theorum be applied to the function f(x)=1/x^2 on the interval [-2, 1]? Explain.
Temka [501]

Answer:

No, it can not be applied.

Step-by-step explanation:

f(x) = 1/x²

f(x) is a polynomial that is not continuous

As,

f(x) = 1/0 is undefines

Secondly, it is not differentiable (i.e. the derivative does not exists on the interval given)

Derivative of this function

f'(x) = (1)x^-2

       = -2x^(-2-1)

       = -2x^(-3)

       = -2/x³

      =  -2/x³

f'(0) = -2/0 is undefined

Thus, mean value theorem can not be applied.

7 0
4 years ago
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