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Lena [83]
3 years ago
6

Yasmine took a survey for her classmates to determine the frequencies of different eye color

Mathematics
1 answer:
Ostrovityanka [42]3 years ago
8 0
What is the question
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Please help will make the BRAINLIEST 50 points !
melomori [17]

Answer:

you given me only 25 point

7 0
3 years ago
What's the perimeter of the wall hanging to the nearest tenth of a centimeter
Temka [501]
You just add the LxW and for area you multiply LxWxLxW
8 0
3 years ago
Can someone help? Please and thankyou<br><br> What is the area of the parallelogram shown?
Thepotemich [5.8K]

Answer:

16

Step-by-step explanation:

Let's give letters as the image. Triangle ABH is a right triangle, so you can apply pythagoream theorem. In particular, it's the classic "3-4-5" triangle (there are some triplets of integers that gives you a right triangle, and 3-4-5 is the most famous). Either way, the height of the triangle is 4.

At that point you can compute the area by multiplying base and height, and the total surface is 16.

4 0
2 years ago
There are 6 fourth-grade students, 8 fifth-grade students, and 10 sixth-grade students on the girls’ soccer team at Truman Eleme
enyata [817]

Answer:

Your answer is

Fourth grade student = 25%

Fifth grade= 33.33%

Sixth grade= 41.67%

Want more answer then follow me, like and MARK MY ANSWER AS BRAINLIST ANSWER.

I need that urgently.

7 0
2 years ago
Urgent:
rjkz [21]

Step-by-step explanation:

<h3>Given:</h3>
  • CM ⊥ AB
  • ∠3 = ∠4
<h3>To prove: </h3>
  • △AMC ≅ △BMC
<h3>Solution:</h3>
  • ∠ACM≅∠BCM - given

<u>CM⊥AB - given, therefore</u>

  • ∠AMC = 90° and ∠BMC = 90°

<u>Then</u>

  • ∠AMC ≅ ∠BMC
  • Side CM is common, therefore is congruent to itself

<u>So we have congruent two angles and a side between them on triangles AMC and BMC:</u>

  • △AMC ≅ △BMC as per ASA congruency theorem
5 0
3 years ago
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