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EleoNora [17]
3 years ago
5

In circle O, diameter AB has been drawn. Locate point C anywhere between A and B (on either side of the diameter). Draw in AC an

d BC . What type of triangle is ACB ? Explain.

Mathematics
1 answer:
-Dominant- [34]3 years ago
8 0

Answer:

Right Triangle.

Step-by-step explanation:

Given AB is a diameter of a circle O.  

Point C is located on the circle's circumference as shown in the attached diagram.

\angle AOB=180^\circ

Theorem: The angle at the centre is twice the angle at the circumference.

The angle formed at centre O by arc AB is \angle AOB

The angle formed at the circumference at point C by arc AB is \angle ACB

By the theorem above:

\angle AOB=2*\angle ACB\\180^\circ=2*\angle ACB\\\angle ACB=90^\circ

Therefore, Triangle ACB is a Right Triangle with the right angle at C.

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4 years ago
*Find the formula of the series 1+2²+3²+4²+....+n²*<br>​
Sophie [7]

The formula is \frac{n(n+1)(2n+1)}{6}

What are series?

In mathematics, we can describe a series as adding infinitely many numbers or quantities to a given starting number or amount.

We will find the formula as shown as below:

Let S=1+2^2+3^2+4^2+................+n^2

We know (n+1)^3=n^3+3n^2+3n+1

(1+1)^3=1^3+3(1)^2+3(1)+1

(2+1)^3=2^3+3(2)^2+3(2)+1

(3+1)^3=3^3+3(3)^2+3(3)+1

.

.

(n+1)^3=n^3+3(n)^2+3(n)+1

On adding

2^3+3^3+4^3......(n+1)^3=(1^3+2^3+3^3+.....+n^3)+3(1^2+2^2+.....+n^2)+3(1+2+3....n)+(1+1+1+....+1)

2^3+3^3+4^3......(n+1)^3-(1^3+2^3+3^3+.....+n^3)=3S+\frac{3n(n+1)}{2}+(1+1+1+....+1)

(n+1)^3-1^3=3S+\frac{3n(n+1)}{2} +n

n^3+3(n)^2+3(n)+1-1=3S+\frac{3n(n+1)}{2} +n

2n^3+6n^2+6n=6S+3n^2+3n+2n

6S=2n^3+3n^2+n

6S=2n^2(n+1)+n(n+1)

6S=(n+1)(2n^2+n)

6S=n(n+1)(2n+1)

S=\frac{n(n+1)(2n+1)}{6}

Hence, the formula is \frac{n(n+1)(2n+1)}{6}

Learn more about Series here:

brainly.com/question/24643676

#SPJ1

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