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coldgirl [10]
2 years ago
12

Which statement correctly names the congruent triangles and justifies the reason for congruence?

Mathematics
2 answers:
Marat540 [252]2 years ago
8 0

Answer:

  C. Triangle BAC is congruent to triangle FDE by AAS

Step-by-step explanation:

BAC names the vertices in the order longest-side, shortest-side. That same order is FDE in the other triangle, eliminating choiced B and D. The triangles are not right triangles, eliminating choice A.

The only viable answer choice is C.

No specific sides are shown as being congruent, but two angles are, so we could claim congruence by ASA or AAS. Answer choice C uses the latter.

goblinko [34]2 years ago
6 0

Answer:

ΔBAC ≅ ΔFDE by AAS

Step-by-step explanation:

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a line whose perpendicular distance from the origin is 4 units and the slope of perpendicular is 2÷3. Find the equation of the l
GrogVix [38]

Answer:

\huge\boxed{y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{2}{3}x+\dfrac{4\sqrt{13}}{3}}

Step-by-step explanation:

The equation of a line:

y=mx+b

We have

m=\dfrac{2}{3}

substitute:

y=\dfrac{2}{3}x+b

The formula of a distance between a point and a line:

General form of a line:

Ax+By+C=0

Point:

(x_0,\ y_0)

Distance:

d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+b^2}}

Convert the equation:

y=\dfrac{2}{3}x+b     |<em>subtract y from both sides</em>

\dfrac{2}{3}x-y+b=0    |<em>multiply both sides by 3</em>

2x-3y+3b=0\to A=2,\ B=-3,\ C=3b

Coordinates of the point:

(0,\ 0)\to x_0=0,\ y_0=0

substitute:

d=4

4=\dfrac{|2\cdot0+(-3)\cdot0+3b|}{\sqrt{2^2+(-3)^2}}\\\\4=\dfrac{|3b|}{\sqrt{4+9}}

4=\dfrac{|3b|}{\sqrt{13}}\qquad|    |<em>multiply both sides by \sqrt{13}</em>

4\sqrt{13}=|3b|\iff3b=-4\sqrt{13}\ \vee\ 3b=4\sqrt{13}   |<em>divide both  sides by 3</em>

b=-\dfrac{4\sqrt{13}}{3}\ \vee\ b=\dfrac{4\sqrt{13}}{3}

Finally:

y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{4\sqrt{13}}{3}

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Answer:

I hope this helps:)

Step-by-step explanation:

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Step-by-step explanation:

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3 years ago
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Answer: a= -18/2

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