Ronald has 5 sets of three yellow beads, if there are 2 blue beads for every set, what is the total amount of blue beads?
Answer:
Third option.
Step-by-step explanation:
You need to cube both sides of the equation. Remember the Power of a power property:
![\sqrt[3]{162x^cy^5}=3x^2y(\sqrt[3]{6y^d})\\\\(\sqrt[3]{162x^cy^5})^3=(3x^2y(\sqrt[3]{6y^d}))^3\\\\162x^cy^5=27x^6y^36y^d](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B162x%5Ecy%5E5%7D%3D3x%5E2y%28%5Csqrt%5B3%5D%7B6y%5Ed%7D%29%5C%5C%5C%5C%28%5Csqrt%5B3%5D%7B162x%5Ecy%5E5%7D%29%5E3%3D%283x%5E2y%28%5Csqrt%5B3%5D%7B6y%5Ed%7D%29%29%5E3%5C%5C%5C%5C162x%5Ecy%5E5%3D27x%5E6y%5E36y%5Ed)
According to the Product of powers property:
Then. simplifying you get:
Now you need to compare the exponents. You can observe that the exponent of "x" on the right side is 6, then the exponent of "x" on the left side must be 6. Therefore:
You can notice that the exponent of "y" on the left side is 5, then the exponent of "x" on the left side must be 5 too. Therefore "d" is:
Answer: The square root of π has attracted attention for almost as long as π itself. When you’re an ancient Greek mathematician studying circles and squares and playing with straightedges and compasses, it’s natural to try to find a circle and a square that have the same area. If you start with the circle and try to find the square, that’s called squaring the circle. If your circle has radius r=1, then its area is πr2 = π, so a square with side-length s has the same area as your circle if s2 = π, that is, if s = sqrt(π). It’s well-known that squaring the circle is impossible in the sense that, if you use the classic Greek tools in the classic Greek manner, you can’t construct a square whose side-length is sqrt(π) (even though you can approximate it as closely as you like); see David Richeson’s new book listed in the References for lots more details about this. But what’s less well-known is that there are (at least!) two other places in mathematics where the square root of π crops up: an infinite product that on its surface makes no sense, and a calculus problem that you can use a surface to solve.
Step-by-step explanation: this is the same paragraph The square root of π has attracted attention for almost as long as π itself. When you’re an ancient Greek mathematician studying circles and squares and playing with straightedges and compasses, it’s natural to try to find a circle and a square that have the same area. If you start with the circle and try to find the square, that’s called squaring the circle. If your circle has radius r=1, then its area is πr2 = π, so a square with side-length s has the same area as your circle if s2 = π, that is, if s = sqrt(π). It’s well-known that squaring the circle is impossible in the sense that, if you use the classic Greek tools in the classic Greek manner, you can’t construct a square whose side-length is sqrt(π) (even though you can approximate it as closely as you like); see David Richeson’s new book listed in the References for lots more details about this. But what’s less well-known is that there are (at least!) two other places in mathematics where the square root of π crops up: an infinite product that on its surface makes no sense, and a calculus problem that you can use a surface to solve.
3 doubled equals 6, so every second multiple of three is also a multiple of six.
Examples:
3 times 4 is 12, and 6 times 2 is 12
3 times 6 is 18, and 6 times 3 is 18
3 times 8 is 24, and 6 times 4 is 24
