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3 years ago

Is 11,13,17 a right triangle

Mathematics

2 answers:

aliya0001 [1]3 years ago

6 0

Answer: no

Step-by-step Explanation: If the sum of the squares of two sides of a triangle is equal to the square of the third side or a² + b² = c², then the triangle is <u>right</u>.

In this problem, we're given a triangle that has

sides with lengths of 11, 13, and 17.

Let's plug these lengths into the pythagorean theorem.

(11)² + (13)² = (17)² or 121 + 169 = 289.

Since 121 + 169 ≠ 289, this is not a right triangle.

11Alexandr11 [23.1K]3 years ago

3 0

Answer:

Step-by-step explanation:

We can use the Pythagorean theorem to check

a^2 +b^2 = c^2

11^2+13^2 = 17^2

121+169 =289

410 does not equal 289

This is not a right triangle

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2 years ago

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3 years ago

Will give brainliest, please help fast

Gekata [30.6K]

Answer:

Converting the equation $x^2-20x+13=0$ into completing the square method we get: $\mathbf{(x-10)^2=87}$

Step-by-step explanation:

we are given quadratic equation: $x^2-20x+13=0$

And we need to convert it into completing the square method.

Completing the square method is of form: $a^2-2ab+b^2=(a-b)^2$

Looking at the given equation $x^2-20x+13=0$

We have a = x

then we have middle term 20x that can be written in form of 2ab So, we have a=x and b=? Multiplying 10 with 2 we get 20 so, we can say that b = 20

So, 20x in form of 2ab can be written as: 2(x)(10)

So, we need to add and subtract (10)^2 on both sides

$x^2-20x+13=0\\x^2-2(x)(10)+(10)^2-(10)^2+13=0\\(x^2-2(x)(10)+(10)^2) \:can\: be\: written\: as\: (x-10)^2 \\(x-10)^2-100+13=0\\(x-10)^2-87=0\\(x-10)^2=87$