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2 years ago

Find the coordinates of centroid of a triangle whose vertices have coordinates (6,2);(- 3, 4); (3,- 9).

Mathematics

1 answer:

Alex777 [14]2 years ago

6 0

$\huge \bf༆ Answer ༄$

Let's solve ~

According to formula, the coordinates of Centroid of a triangle are ;

$\sf x = \dfrac{x_1 + x_2 + x_3 }{3}$

$\sf y = \dfrac{y_1 + y_2 + y_3 }{3}$

So, plug the coordinates of its vertices in the formula to find the required solution !

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:x = \dfrac{6 + ( - 3) + 3}{3}$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:y = \dfrac{2+ 4 + (- 9) }{3}$

Further solving ~

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:x = \dfrac{6}{3}$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:y = \dfrac{ - 3}{3}$

Therefore, the required coordinates are ~

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:x = 2$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:y = - 1$

Correct choice is ;

$\overbrace{ \underbrace{\underline{ \boxed{ \sf (2 ,-1)}}}}$

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3 years ago

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2 years ago

What value of x is in the solution set of 3(x – 4) ≥ 5x + 2?

Aleks04 [339]

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Subtract 2

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3 years ago

Read 2 more answers

Find a formula for dy/dx if sin x + cos y + sec(xy) = 251

Lena [83]

Answer:

$\displaystyle \frac{dy}{dx} = \frac{-cos(x) - ysec(xy)tan(xy)}{-sin(y) + xsec(xy)tan(xy)}$

General Formulas and Concepts:

Pre-Algebra

Distributive Property

Algebra I

Factoring

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Trig Differentiation

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Implicit Differentiation

Step-by-step explanation:

Step 1: Define

Identify

sin(x) + cos(y) + sec(xy) = 251

Step 2: Differentiate

[Implicit Differentiation] Trig Differentiation [Chain Rule]: $\displaystyle cos(x) - sin(y)\frac{dy}{dx} + sec(xy)tan(xy) \cdot (y + x\frac{dy}{dx}) = 0$
[Subtraction Property of Equality] Isolate $\displaystyle \frac{dy}{dx}$ terms: $\displaystyle -sin(y)\frac{dy}{dx} + sec(xy)tan(xy) \cdot (y + x\frac{dy}{dx}) = -cos(x)$
[Distributive Property] Distribute sec(xy)tan(xy): $\displaystyle -sin(y)\frac{dy}{dx} + ysec(xy)tan(xy) + xsec(xy)tan(xy)\frac{dy}{dx} = -cos(x)$
[Subtraction Property of Equality] Isolate $\displaystyle \frac{dy}{dx}$ terms: $\displaystyle -sin(y)\frac{dy}{dx} + xsec(xy)tan(xy)\frac{dy}{dx} = -cos(x) - ysec(xy)tan(xy)$
Factor out $\displaystyle \frac{dy}{dx}$ : $\displaystyle \frac{dy}{dx}[-sin(y) + xsec(xy)tan(xy)] = -cos(x) - ysec(xy)tan(xy)$
[Division Property of Equality] Isolate $\displaystyle \frac{dy}{dx}$ : $\displaystyle \frac{dy}{dx} = \frac{-cos(x) - ysec(xy)tan(xy)}{-sin(y) + xsec(xy)tan(xy)}$