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zhuklara [117]
3 years ago
9

Help pls I need it I will give brainiest ​

Mathematics
1 answer:
noname [10]3 years ago
5 0
If x=4 y=1 as shown in the pic
The answer will be 1
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Square has side lengths of 13 units. Point lies in the interior of the square such that units and units. What is the distance fr
romanna [79]

The distance from E to side AD is 25/13.

<h3>What is a distance?</h3>
  • The length of the line connecting two places is the distance between them.
  • If the two points are on the same horizontal or vertical line, the distance can be calculated by subtracting the non-identical values.

To find what is the distance from E to side AD:

  • If you draw a diagram, you'll see that triangle AEB is a right triangle with lengths 5, 12, and 13.
  • Let's call F the point where E meets side AD, so the problem is to find the length of EF.
  • By Angle-Angle Similarity, triangle AFE is similar to triangle BEA. (the right angles are congruent, and both angle FAE and ABE are complementary to angle BAE)
  • Since they're similar, the ratios of their side lengths are the same.
  • EF/EA = EA/AB (they're corresponding side lengths of similar triangles).

Substitute them with known lengths:

  • EF/5 = 5/13
  • EF = 5 × (5/13) = 25/13

Therefore, the distance from E to side AD is 25/13.

Know more about distance here:

brainly.com/question/2854969

#SPJ4

The correct answer is given below:
Square ABCD has side lengths of 13 units. Point E lies in the interior of the square such that AE=5 units and BE=12 units. What is the distance from E to side AD? Express your answer as a mixed number.

8 0
1 year ago
Last one for this test
Oksanka [162]

Answer:

1) 0

2) 0

3) 0

Explanation:

6 0
3 years ago
Nind reeds
Usimov [2.4K]

Answer:

42 gallons.

Step-by-step explanation:

count every inch *2 and calculate gallons

3 0
2 years ago
PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!<br><br> Subract and simplify.
pantera1 [17]

Answer: C

<u>Step-by-step explanation:</u>

  \dfrac{x+1}{x-5} - \dfrac{x-2}{x+3}

= \dfrac{x+1}{x-5}(\dfrac{x+3}{x+3}) - \dfrac{x-2}{x+3}(\dfrac{x-5}{x-5})

= \dfrac{x^{2}+4x+3}{(x-5)(x+3)} - \dfrac{x^{2}-7x+10}{(x-5)(x+3)}

= \dfrac{x^{2}+4x+3-(x^{2}-7x+10)}{(x-5)(x+3)}

= \dfrac{11x-7}{(x-5)(x+3)}

5 0
3 years ago
Please help please please please
Readme [11.4K]
It is a low posabilaty
8 0
3 years ago
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