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Misha Larkins [42]
2 years ago
12

Perimeter of a equilateral triangle is 27cm. Find the length of each sides.

Mathematics
1 answer:
Mademuasel [1]2 years ago
7 0

Answer:

EQUILATERAL TRIANGLE = 3/27

\frac{27}{3 = } 9

<h2>SO THE ANS IS 9 .</h2>

I HOPE IT IS HELPFUL

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Please help due today
otez555 [7]
The answer is c
12.4 feet by 9.5 feet
4 0
2 years ago
the length of an arc of a circle is 1/5 of its circumference. if the area of the circle is346.5cm²,find the angle subtended by t
dexar [7]

Answer:

72°

Step-by-step explanation:

From the question,

Area of the circle = πr²

A = πr²................. Equation 1

Where r = radius of the circle.

⇒ r = √(A/π)............. Equation 2

Given: A = 346.5 cm², π = 3.14

r = √(346.5/3.14)

r = √(110.35)

r = 10.5 cm.

Therefore,

circumference of the circle = 2πr = 2×3.14×10.5

circumference = 65.94 m

If the length of the arc(s) is 1/5 of its circumference.

Therefore, length of arc (s) = 13.188

⇒ length of arc/circumference = 13.188/65.94 =  1/5

s/2πr = θ/360

Where θ = angle substends at the center of the circle

1/5 = θ/360

θ = 360/5

θ = 72°

3 0
2 years ago
How Do I do proofs for proving triangles?
MA_775_DIABLO [31]
You must have been taught postulates and theorems that allow you to prove triangles congruent, such as SSS, SAS, ASA, etc. Look at the given information of a proof, and see how from the given information, using definitions, postulates, and theorems you have already learned, you can show pairs of corresponding sides and angles to be congruent that will fit into the above methods. Then use one of the methods to prove the triangles congruent.
4 0
3 years ago
Para reunir dinero para su gira de estudios , los alumnos de un curso deciden vender números de una rifa que se encuentran numer
QveST [7]

Respuesta:

0.53

Explicación:

Para calcular la posibilidad del evento A: "ganar la rifa comprando todos los números múltiplos de 3 o 5", debemos usar la siguiente fórmula.

P(A) = casos favorables / casos posibles

Evaluemos primero todos los casos que son múltiplos de 3, entre 1 y 100: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99. En total son 33.

Ahora, evaluemos todos los casos que son múltiplos de 3, entre 1 y 100: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100. En total son 20.

El número total de casos favorables es 33 + 20 = 53.

El número de casos posibles es el total de números de 1 a 100, es decir 100.

Luego P(A) = 53/100 = 0.53.

7 0
2 years ago
How is 17 + 26 different from 36 + 53? Explain ( this is regrouping lesson)
xxMikexx [17]
17+26 is different from 36+53 by if you add 17+26 the sum would be 43. And if you add 36+53 the sum would also be 89. So that means the way how they are  different is that the 17+26 and 36+53 has different kinds of sums. They have different kinds of sums because you are different numbers.
 Hope this helped and if you need more help then just message me instead :)
6 0
3 years ago
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