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Otrada [13]
2 years ago
12

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

<u>Pre-Algebra</u>

Distributive Property

<u>Algebra I</u>

  • Factoring

<u>Calculus</u>

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:

<u>Step 1: Define</u>

<em>Identify</em>

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

<u>Step 2: Differentiate</u>

  1. [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                      
  2. [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)
  3. [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)
  4. [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)
  5. Factor out  \displaystyle \frac{dy}{dx}:                                                                                                   \displaystyle \frac{dy}{dx}[-sin(y) + xsec(xy)tan(xy)] = -cos(x) - ysec(xy)tan(xy)
  6. [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)}

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Implicit Differentiation

Book: College Calculus 10e

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