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Dennis_Churaev [7]
3 years ago
16

The length of ZX is 2 units. What is the perimeter of triangle XYZ? 5 + + 2 units 5 + 3 units 5 + + 2 units 10 + 2 units

Mathematics
1 answer:
Shtirlitz [24]3 years ago
7 0

Answer:

B) 5 + 3√5 units

Step-by-step explanation:

The length of ZX is 2√5 units. What is the perimeter of triangle XYZ?  

A) 5 +√3 + 2 √5 units  

B) 5 + 3√5 units  

C) 5 + √6 + 2√5 units  

D) 10 + 2√5 units

From the diagram attached, point X is at (-1, 4), Y(3, 1), Z(1, 0).

The distance between two point

O(x_1,y_1)\ and\ A(x_2,y_2)\ is\ given\ as:\\\\OA=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}

The lengths of the sides of the triangle are:

|XY| = \sqrt{(3-(-1))^2+(1-4)^2}=\sqrt{25} =5\ unit\\ \\|XZ|= \sqrt{(1-(-1))^2+(0-4)^2}=\sqrt{20} =2\sqrt{5} \ unit\\\\|YZ|= \sqrt{(1-3)^2+(0-1)^2}=\sqrt{5} \ unit

The perimeter of the triangle is the sum of all the sides, i.e.

Perimeter = |XY| + |YZ| + |XZ| = 5 + 2√5 + √5 = 5 + 3√5  

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\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}

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\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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