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2 years ago

What else would need to be congruent to show that ABC is congruent to DEF by ASA?

Mathematics

1 answer:

cestrela7 [59]2 years ago

5 0

Answer:

∠ A ≅ ∠ D

Step-by-step explanation:

We have to prove Δ ABC and Δ DEF are congruent or Δ ABC ≅ Δ DEF applying the Angle-Side-Angle or ASA rule.

In order to maintain the sequence, in ASA rule we have to make congruent two angles and one side in between them.

Therefore, if ∠ B and ∠ E are proved to be congruent and segment AB and segment DE are proved to be congruent, then we have to consider another two angles which must be ∠ A and ∠ D to be congruent.

Therefore, option D is correct. (Answer)

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Answer:

y = -2x + 1

Step-by-step explanation:

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A mile is equal to 5280 feet.if the highway department places a reflector every 25 feet how many reflectors will there be in 1 m

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2 years ago

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<img src="https://tex.z-dn.net/?f=0%20%3D%20%20%20-%204.9t%20%7B%7D%5E%7B2%7D%20%20%2B%2027t%20%2B%202.4" id="TexFormula1" title

marta [7]

$0 = - 4.9t {}^{2} + 27t + 2.4$
Convert decimals to fractions

$0 = - \frac{49}{10} t {}^{2} + 27t + \frac{12}{5}$
Multiply both sides by 10

$0 = - 49t {}^{2} + 270t + 24$
Move expression to the left.

$0 + 49t {}^{2} - 270t - 24 = 0$
Remove zero.

$49t {}^{2} - 270t - 24 = 0$ Solve the quadratic equation.

$t = \frac{ - ( - 270) + - \sqrt{( - 270) {}^{2} - 4 \times 49 \times ( - 24) } }{2 \times 49}$
Remove parenthesis and calculate.

$t = \frac{270 + - \sqrt{72900 + 4704} }{98}$
Add the numbers

$t = \frac{270 + - 77604}{98}$
Separate the solutions

$t = \frac{270 + \sqrt{77604} }{98} \\ t = \frac{270 - \sqrt{77604} }{98}$
The final solution are
$t1 = \frac{270 - \sqrt{77604} }{98} , t2 = \frac{270 + \sqrt{77604} }{98}$

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3 years ago

A surveyor is conducting measures to see how wide a road is

drek231 [11]

Answer:

D) 45 ft

Step-by-step explanation:

The two triangles are shown below.

Given:

BC = 60 ft, CD = 24 ft and DE = 18 ft.

Since, the two triangles are similar, their corresponding sides are in proportion.

So, $\frac{AB}{DE}=\frac{BC}{CD}=\frac{AC}{CE}$