4. To determine if a triangle is a right triangle, given that you know the length of its sides, you have to check if its lengths follow the Pythagorean theorem.
This theorem states that the square of the hypothenuse (c) is equal to the sum of the squares of the legs of the triangle (a and b), following the expression:

The triangle is:
We have to check that a²+ b² is equal to c².
The square of the hypothenuse is:

The sum of the squares of the legs of the triangle is:

As you can see, the sum of the squares of the legs of the triangle is 100, which is the same as the square of the hypothenuse. The triangle follows the Pythagorean theorem and can be considered a right triangle.
Answer:
4(3n+2) or 12n+8
Step-by-step explanation:
Given expression is:

The numerator of the fraction will be multiplied with 9n^2- 4
So, Multiplication will give us:

We can simplify the expression before multiplication.
The numerator will be broken down using the formula:
![a^2 - b^2 = (a+b)(a-b)\\So,\\= \frac{8[(3n)^2 - (2)^2]}{6n-4}\\ = \frac{8(3n-2)(3n+2)}{6n-4}](https://tex.z-dn.net/?f=a%5E2%20-%20b%5E2%20%3D%20%28a%2Bb%29%28a-b%29%5C%5CSo%2C%5C%5C%3D%20%5Cfrac%7B8%5B%283n%29%5E2%20-%20%282%29%5E2%5D%7D%7B6n-4%7D%5C%5C%20%3D%20%5Cfrac%7B8%283n-2%29%283n%2B2%29%7D%7B6n-4%7D)
We can take 2 as common factor from denominator

Hence the product is 4(3n+2) or 12n+8 ..
Answer:
x=137
y=525-137
y=388
Step-by-step explanation:
Let the student tickets be x
Let the geral Admission tickets be y
x+y=525
y=525-x
4x+6y=2876 (subsitute for y)
4x+6(525-x)=2876
4x+3150-6x=2876
-2x=-274
x=137
y=525-137
y=388
Hence, about 137 childern tickets were sold and 388 gernal admission tickets were sold.
Paul.
<h3>Explanation:</h3>
GCF: the greatest common factor of numerator and denominator is a factor that can be removed to reduce the fraction.
<em>Example</em>
The numerator and denominator of 6/8 have GCF of 2:
6/8 = (2·3)/(2·4)
The fraction can be reduced by canceling those factors.
(2·3)/(2·4) = (2/2)·(3/4) = 1·(3/4) = 3/4
___
LCM: the least common multiple of the denominators is suitable as a common denominator. Addition and subtraction are easily performed on the numerators when the denominator is common.
<em>Example</em>
The fractions 2/3 and 1/5 can be added using a common denominator of LCM(3, 5) = 15.
2/3 + 1/5 = 10/15 + 3/15 = (10+3)/15 = 13/15