A ratio of 5:7 means out of 12 books, 5 are in one category and 7 are in another. (The “category” isn’t specified in this problem.)
So here we’re STARTING with 36 books; we want to find a ratio that is equal to 5:7 and adds up to 36. Well 36 = 3(12), and so if we multiply each number in the ratio by 3 as well, we might get what we want. Doing that gives us 15:21, which adds up to 36 and is equal to 5:7. Boom, that’s the answer.
P.S. Here’s a question you can think about: why do ratios stay the same when you multiply each number by the same amount? (For example, why is 3:1 the same as 6:2?)
Answer:
![4x^{3} y^{2} (\sqrt[3]{4 x y})](https://tex.z-dn.net/?f=4x%5E%7B3%7D%20y%5E%7B2%7D%20%28%5Csqrt%5B3%5D%7B4%20x%20y%7D%29)
Step-by-step explanation:
Another complex expression, let's simplify it step by step...
We'll start by re-writing 256 as 4^4
![\sqrt[3]{256 x^{10} y^{7} } = \sqrt[3]{4^{4} x^{10} y^{7} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B256%20x%5E%7B10%7D%20y%5E%7B7%7D%20%7D%20%3D%20%5Csqrt%5B3%5D%7B4%5E%7B4%7D%20x%5E%7B10%7D%20y%5E%7B7%7D%20%7D)
Then we'll extract the 4 from the cubic root. We will then subtract 3 from the exponent (4) to get to a simple 4 inside, and a 4 outside.
![\sqrt[3]{4^{4} x^{10} y^{7} } = 4 \sqrt[3]{4 x^{10} y^{7} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B4%5E%7B4%7D%20x%5E%7B10%7D%20y%5E%7B7%7D%20%7D%20%3D%204%20%5Csqrt%5B3%5D%7B4%20x%5E%7B10%7D%20y%5E%7B7%7D%20%7D)
Now, we have x^10, so if we divide the exponent by the root factor, we get 10/3 = 3 1/3, which means we will extract x^9 that will become x^3 outside and x will remain inside.
![4 \sqrt[3]{4 x^{10} y^{7} } = 4x^{3} \sqrt[3]{4 x y^{7} }](https://tex.z-dn.net/?f=4%20%5Csqrt%5B3%5D%7B4%20x%5E%7B10%7D%20y%5E%7B7%7D%20%7D%20%3D%204x%5E%7B3%7D%20%5Csqrt%5B3%5D%7B4%20x%20y%5E%7B7%7D%20%7D)
For the y's we have y^7 inside the cubic root, that means the true exponent is y^(7/3)... so we can extract y^2 and 1 y will remain inside.
![4x^{3} \sqrt[3]{4 x y^{7} } = 4x^{3} y^{2} \sqrt[3]{4 x y}](https://tex.z-dn.net/?f=4x%5E%7B3%7D%20%5Csqrt%5B3%5D%7B4%20x%20y%5E%7B7%7D%20%7D%20%3D%204x%5E%7B3%7D%20y%5E%7B2%7D%20%5Csqrt%5B3%5D%7B4%20x%20y%7D)
The answer is then:
![4x^{3} y^{2} \sqrt[3]{4 x y} = 4x^{3} y^{2} (\sqrt[3]{4 x y})](https://tex.z-dn.net/?f=4x%5E%7B3%7D%20y%5E%7B2%7D%20%5Csqrt%5B3%5D%7B4%20x%20y%7D%20%3D%204x%5E%7B3%7D%20y%5E%7B2%7D%20%28%5Csqrt%5B3%5D%7B4%20x%20y%7D%29)
We know that
X²+y²=9 -------> X²+y²=3²
is the equation of a circle with center (0,0) and radius r=3 units
so
<span>the translation of four units to the right and three units down is equals to move the center (0,0)--------> (0+4,0-3)------> (4.-3)
the new center of the circle is (4,-3)
the new equation is
(x-4)</span>²+(y+3)²=3²
see the attached figure
Answer:
y = 2
Step-by-step explanation:
x= -1+y
2x-y=0
2(-1+y)-y=0
-2+2y-y=0
-2+y=0
-2+2+y=0+2
y=2
hope it's helpful ❤❤❤❤
THANK YOU.
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Answer:
Ryan earned $52 Mike earned $109 and Stormy earned $94
Step-by-step explanation:
x+2×+5+2x-10=255
5x-5=255
5x=260
x=52