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insens350 [35]
2 years ago
5

Given that T is the centroid of △DEF and FT=12

Mathematics
1 answer:
dlinn [17]2 years ago
6 0

The centroid of a triangle divides the median of the triangle into 1 : 2

The measure of FQ is 18, while the measure of TQ is 6

Because point T is the centroid, then we have the following ratio

\mathbf{TQ : FT =1 : 2}

Where FT = 12.

Substitute 12 for FT in the above ratio

\mathbf{TQ : 12 =1 : 2}

Express as fraction

\mathbf{\frac{TQ }{ 12} =\frac{1 }{ 2}}

Multiply both sides by 12

\mathbf{TQ =\frac{1 }{ 2} \times 12}

This gives

\mathbf{TQ =\frac{1 2}{ 2}}

Divide 12 by 2

\mathbf{TQ =6}

The measure of FQ is calculated using:

\mathbf{FQ = FT + TQ}

Substitute 12 for FT, and 6 for TQ

\mathbf{FQ = 12 + 6}

Add 12 and 6

\mathbf{FQ = 18}

Hence, the measure of FQ is 18, while the measure of TQ is 6

Read more about centroids at:

brainly.com/question/11891965

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Answer:

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Step-by-step explanation:

a) y = 1/3x, so each value satisfies it. This is true.

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What is the slope of the line that passes through the points (1, -6) and (-8, 9) ?Write your answer in simplest form.
Dmitry [639]

Answer:

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Step-by-step explanation:

The slope formula is rise/run.

If you graph this, the two points make a line that slopes downwards. The higher point is in the negative axis, which means that the run will be negative.

you rise up 15 points from the first point (1,-6) to get to the same level as the second point. (-8,9). This is your rise.

You go over to the other point, this is your run. The run is -9.

15/-9

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Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that
kherson [118]

Answer:

The answer is

f(x) = {\displaystyle  8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }

Step-by-step explanation:

Remember that Taylor says that

f(x) = {\displaystyle \sum\limits_{k=0}^{\infty} \frac{f^{(k)}(a) }{k!}(x-a)^k }

For this case

f^{(0)} (-2) = 8(-2)-3(-2)^3 = 8\\f^{(1)} (-2) = 8-3(3)(-2)^2 = -28\\f^{(2)} (-2) = -3(3)2(-2) = -36\\f^{(2)} (-2) = -3(3)2 = -18

f(x) = {\displaystyle  8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }

5 0
3 years ago
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