32*24=768
16(8*6)=16(48)=768
64(4*3)=64(12)=768
you can do 5 12 times
commutative property 9X3
It is 113/93. You can simplify the ratio of required.
To find the product of 17 and a variable "s," you simply multiply 17 by s. The result of 17 times s is written as 17s, which represents the product of 17 and the value of s. This operation involves scalar multiplication in algebra, where a number is multiplied by a variable.
You could use it because it shows that its just 7 times 8 flipped!
No, not necessary. To simplify something is to make the expressions "clean and clearer". For instance: 2 / 4 = 1 / 2 [simplified form]
To find the product of 0.8 and 0.07, you can simply multiply the two numbers together. So, 0.8 times 0.07 equals 0.056. This can also be expressed as 56/1000 or 7/125.
It means find the value of the expression. It cannot be simplified in the way that algebraic expressions usually are.
The product is: 45 times 39 = 1755
Yes - to find the product of a set of numbers is to multiply them together.
To add and subtract algebraic expressions the simple rule of like terms applies. In your homework that asks for the expression represents the perimeter in units of this trapezoid you will need to find the like terms and simplify.
first use 5 times 6 = 30 than add 6 times 1 = 6 so the product of 6 times 6 is 36
you do 450 over 100 times 870 over 1 the answer you get just simplify. :)
you times the two numbers together for example find the product of 5 and 6 =5x6 =30 Product means 'times' (X) Example the product of 5 and 2 is 10 Because 5 X 2 = 10.
we can check it by 2 times
75 multiplied by 5 equals 375. This calculation involves multiplying the two numbers together to find the product. To arrive at this answer, you can add 75 to itself five times, or use the multiplication operation to simplify the process.
Sample Answer: To verify, a number needs to be substituted for x in both expressions. Use order of operations to simplify and find the value. The value needs to be the same for both expressions to prove equivalence.