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2 years ago

Least common denominator of 3/5 and 11/20

1 answer:

6 0

Solution:

Rewriting input as fractions if necessary:
3/5, 11/20

For the denominators (5, 20) the least common multiple (LCM) is 20.

LCM(5, 20)

Therefore, the least common denominator (LCD) is 20.

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Step-by-step explanation:

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2 years ago

I need help please, I will give brainliest to the first person that answers. Also extra points.

Papessa [141]

Answer:

a + c = 7

9a + 4c = $43

Step-by-step explanation:

There're 7 tickets which were bough in total. Two different types of tickets, one which represented children, the other for adults. The adult ticket is represented by a and is 9 dollars. The children's ticket is represented by c and is 4 dollars.

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2 years ago

Find the exact area of an equilateral triangle if the length of one side is 40

VARVARA [1.3K]

Equilateral triangle is a triangle with the 3 sides of the same length, and the 3 angles with the same measures/degree.
The formula to find the area of a triangle is base x height/2 , you do have the base (40), but you need the height. To find the height of an equilateral triangle you use the formula: side/2 x radical3, this way u'll find the height and therefor can calculate the area of the triangle.
Hope that helped :)

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2 years ago

. Your class is collecting bottled water for a service project. The goal is to collect 300 bottles of water. On the first day, M

kobusy [5.1K]

6x12=72
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3 years ago

Read 2 more answers

Use the Polynomial Identity below to help you create a list of 10 Pythagorean Triples:

Stolb23 [73]

Answer:

(3,4,5)

(6,8,10)

(5,12,13)

(8,15,17)

(12,16,20)

(7,24,25)

(10,24,26)

(20,21,29)

(16,30,34)

(9,40,41)

Just choose 2 numbers from {1,2,3,4,5,6,7,8,...} and make sure the one you input for x is larger.

Post the three in the comments and I will check them for you.

Step-by-step explanation:

We need to choose 2 positive integers for x and y where x>y.

Positive integers are {1,2,3,4,5,6,7,.....}.

I'm going to start with (x,y)=(2,1).

x=2 and y=1.

$(2^2+1^2)^2=(2^2-1^2)^2+(2\cdot2\cdot1)^2$

$(4+1)^2=(4-1)^2+(4)^2$

$(5)^2=(3)^2+(4)^2$

So one Pythagorean Triple is (3,4,5).

I'm going to choose (x,y)=(3,1).

x=3 and y=1.

$(3^2+1^2)^2=(3^2-1^2)^2+(2\cdot3\cdot1)^2$

$(9+1)^2=(9-1)^2+(6)^2$

$(10)^2=(8)^2+(6)^2$

So another Pythagorean Triple is (6,8,10).

I'm going to choose (x,y)=(3,2).

x=3 and y=2.

$(3^2+2^2)^2=(3^2-2^2)^2+(2\cdot3\cdot2)^2$

$(9+4)^2=(9-4)^2+(12)^2$

$(13)^2=(5)^2+(12)^2$

So another is (5,12,13).

I'm going to choose (x,y)=(4,1).

$(4^2+1^2)^2=(4^2-1^2)^2+(2\cdot4\cdot1)^2$

$(16+1)^2=(16-1)^2+(8)^2$

$(17)^2=(15)^2+(8)^2$

Another is (8,15,17).

I'm going to choose (x,y)=(4,2).

$(4^2+2^2)^2=(4^2-2^2)^2+(2\cdot4\cdot2)^2$

$(16+4)^2=(16-4)^2+(16)^2$

$(20)^2=(12)^2+(16)^2$

We have another which is (12,16,20).

I'm going to choose (x,y)=(4,3).

$(4^2+3^2)^2=(4^2-3^2)^2+(2\cdot4\cdot3)^2$

$(16+9)^2=(16-9)^2+(24)^2$

$(25)^2=(7)^2+(24)^2$

We have another is (7,24,25).

You are just choosing numbers from the positive integer set {1,2,3,4,... } and making sure the number you plug in for x is higher than the number for y.

I will do one more.

Let's choose (x,y)=(5,1).

$(5^2+1^2)^2=(5^2-1^2)^2+(2\cdot5\cdot1)^2$