First, I'm going to separate factor the expression inside of the square root.
sqrt[ (2/18) * (x^5) ]
sqrt[ (1/9) * (x^5) ]
We can take the square root of 1/9 easily, because 1 and 9 are both perfect squares. The square root of 1/9 is 1/3.
Looking at the x^5, we can separate it into x^2 * x^2 * x^1. The square root of x^2 is x. So, we now also have an x^2 on the outside because there are two x^2s in our expanded form.
ANSWER: (x^2 * sqrt(x)) / 3
(Option 1)
Hope this helps!
Pick one coordinate point on figure 1 (-4,5)
First rotate 90 degree clockwise rule (x,y) -->(y , -x)
So (-4,5) rotate 90 degree clockwise will become (5, 4)
Then reflection over x a-xis, rule (x,y) -->(x,-y)
So (5, 4) reflection over x a-xis will be (5, -4)
Answer: first option
90 clockwise rotation around the origin, then a reflection across x - axis.
Hope it helps
Given:
27y and 54y³
To find:
The highest common factor (HCF) using prime factorization.
Solution:
We have,
27y and 54y³
Using the prime factorization, we get


Now, the common prime factors are 3, 3, 3 and y. So,


Therefore, the highest common factor of the given terms is 27y.
The closest answer I got was 32
Check the picture below.
let's notice that the angle at K is an inscribed angle with an intercepted arc
![\bf \stackrel{\textit{using the inscribed angle theorem}}{K=\cfrac{\widehat{LI}+\widehat{IJ}}{2}}\implies 9x+1=\cfrac{(10x-1)+59}{2} \\\\\\ 9x+1=\cfrac{10x+58}{2}\implies 18x+2=10x+58\implies 8x+2=58 \\\\\\ 8x=56\implies x=\cfrac{56}{8}\implies x=7 \\\\[-0.35em] ~\dotfill\\\\ K=9x+1\implies K=9(7)+1\implies \boxed{K=64}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Busing%20the%20inscribed%20angle%20theorem%7D%7D%7BK%3D%5Ccfrac%7B%5Cwidehat%7BLI%7D%2B%5Cwidehat%7BIJ%7D%7D%7B2%7D%7D%5Cimplies%209x%2B1%3D%5Ccfrac%7B%2810x-1%29%2B59%7D%7B2%7D%20%5C%5C%5C%5C%5C%5C%209x%2B1%3D%5Ccfrac%7B10x%2B58%7D%7B2%7D%5Cimplies%2018x%2B2%3D10x%2B58%5Cimplies%208x%2B2%3D58%20%5C%5C%5C%5C%5C%5C%208x%3D56%5Cimplies%20x%3D%5Ccfrac%7B56%7D%7B8%7D%5Cimplies%20x%3D7%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20K%3D9x%2B1%5Cimplies%20K%3D9%287%29%2B1%5Cimplies%20%5Cboxed%7BK%3D64%7D)
now, let's notice something again, the angle at L is also an inscribed angle, intercepting and arc of 97 + 59 = 156, so then, by the inscribed angle theorem,
∡L is half that, or 78°.
now, let's take a look at the picture down below, to the inscribed quadrilateral conjecture, since ∡J and ∡I are both supplementary angles, then
∡I = 180 - 64 = 116°.
∡J = 180 - 78 = 102°.