3/5 of the visitors to the museum come during the week, and 4/10 come during the weekend. When does the museum get more visitors, during the week or during the weekend.
We know that 3/5 of the visitors to the museum come during the week, and that 4/10 come during the weekend. Since we want to know when the museum gets more visitors, we need to compare 3/5 to 4/10.
Notice that 3/5 and 4/10 are unlike fractions so our first step is to find a common denominator. The common denominator of 5 and 10 is the least common multiple of 5 and 10 or 10.
To get a denominator of 10 for our first fraction 3/5, we must multiply both the numerator and denominator by 2 to get 6/10.
Notice that 4/10 already has 10 in the denominator so it stays the same. Now we have like fractions. To compare like fractions, we simply compare the numerators or the top numbers. Since 6 is greater than 4, we know that 6/10 must be greater than 4/10. Since 6/10 represents the number of visitors that come to the museum during the week, we know that the museum gets more visitors during the week.
side
Answer:
your answer depends on what sides are given.
For example, lets say that you now the lengths of two sides, these being 56 and 24. ( these are just random numbers BTW) to find the third you would add the two lengths that you know and then minus them from 180. So for the example i just typed you would do 56 + 24 first which equals 80, then would do 180-80 which equals 100.
To check your answer, you could do this:
DO THE ABOVE AGAIN CAREFULLY (lol I don't usually check so.. idk)
math is my worst subject so.... hope this helped you.
Answer:
Use the Pythagorean Theorem: a² + b² = c².
Step-by-step explanation:
For any given right triangle the sum of the measures of the sides squared is equal to the hypotenuse squared. The formula is call the Pythagorean Theorem and is written: a² + b² = c², where 'a' and 'b' are the legs of the triangle and 'c' is the hypotenuse.
Depending on which side of the triangle they give you, you can plug the other two side lengths into the equation to solve for the third.