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Archy [21]
2 years ago
7

∆ABC is translated 6 units up and 3 units left to create ∆A'B'C'.

Mathematics
1 answer:
Rama09 [41]2 years ago
5 0

A triangle is a three-edged polygon with three vertices. It is a fundamental form in geometry. The vertex B is at (1, 5), then vertex B' is at (-2, 11).

<h3>What is a triangle?</h3>

A triangle is a three-edged polygon with three vertices. It is a fundamental form in geometry. The sum of all the angle of a triangle is always equal to 180°.

∆ABC is translated 6 units up and 3 units left to create ∆A'B'C'.

The vertex A is at (-1, 2), then vertex A' is at,

A' = (-1-3, 2+6) = (-4, 8)

The vertex B is at (1, 5), then vertex B' is at,

B' = (1-3, 5+6) = (-2, 11)

Learn more about Triangle:

brainly.com/question/2773823

#SPJ1

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4 0
3 years ago
Lin sold 4 more shirts than Greg. Fran sold 3 times as many shirts as Lin. In total the three sold 51 shirts. Which represents t
Alex17521 [72]
 <span>Your Answer: D 

Why.... 

You can make equations out of the information 
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L = 4 + G ---"Lin sold 4 more shirts than Greg" 
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Use F + G + L = 51 
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(3L) + G + (4+G) = 51 

You still have a variable besides G in there... you can use the L= 4+G and substitute again so that there are only G's 

( 3(4+G) ) + G + (4+G) = 51 ---- SIMPLIFY :D 
( 12 + 3G ) + G + 4 + G = 51 
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16 + 5G = 51 </span>
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4 years ago
Calculate the total area of the shaded region.
LUCKY_DIMON [66]

so hmmm seemingly the graphs meet at -2 and +2 and 0, let's check

\stackrel{f(x)}{2x^3-x^2-5x}~~ = ~~\stackrel{g(x)}{-x^2+3x}\implies 2x^3-5x=3x\implies 2x^3-8x=0 \\\\\\ 2x(x^2-4)=0\implies x^2=4\implies x=\pm\sqrt{4}\implies x= \begin{cases} 0\\ \pm 2 \end{cases}

so f(x) = g(x) at those points, so let's take the integral of the top - bottom functions for both intervals, namely f(x) - g(x) from -2 to 0 and g(x) - f(x) from 0 to +2.

\stackrel{f(x)}{2x^3-x^2-5x}~~ - ~~[\stackrel{g(x)}{-x^2+3x}]\implies 2x^3-x^2-5x+x^2-3x \\\\\\ 2x^3-8x\implies 2(x^3-4x)\implies \displaystyle 2\int\limits_{-2}^{0} (x^3-4x)dx \implies 2\left[ \cfrac{x^4}{4}-2x^2 \right]_{-2}^{0}\implies \boxed{8} \\\\[-0.35em] ~\dotfill

\stackrel{g(x)}{-x^2+3x}~~ - ~~[\stackrel{f(x)}{2x^3-x^2-5x}]\implies -x^2+3x-2x^3+x^2+5x \\\\\\ -2x^3+8x\implies 2(-x^3+4x) \\\\\\ \displaystyle 2\int\limits_{0}^{2} (-x^3+4x)dx \implies 2\left[ -\cfrac{x^4}{4}+2x^2 \right]_{0}^{2}\implies \boxed{8} ~\hfill \boxed{\stackrel{\textit{total area}}{8~~ + ~~8~~ = ~~16}}

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