Because this is an equiangular triangle, we can set any two sides equal to each-other to solve for x, because all sides are the same length.
5x - 22 = 4x - 10
Subtract 4x from both sides.
x - 22 = -10
Add 22 to both sides.
x = 12
<h3>The value of x is equal to 12.</h3>
Answer: 16
Step-by-step explanation: 27-11=16
The decrease in the value of the toy is $9.25 if you subtract $ 0.75 from $10.00 then you get $9.25 so it's $9.25 cheaper
Answer:
![\dfrac{\sqrt[3]{95^2}}{17\cdot95^4}=\dfrac{\sqrt[3]{9\,025}}{1\,384\,660\,625}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B95%5E2%7D%7D%7B17%5Ccdot95%5E4%7D%3D%5Cdfrac%7B%5Csqrt%5B3%5D%7B9%5C%2C025%7D%7D%7B1%5C%2C384%5C%2C660%5C%2C625%7D)
Step-by-step explanation:
The applicable rules of exponents are ...
(ab)^c = (a^c)(b^c)
(a^b)/(a^c) = a^(b-c)
__
![\dfrac{190^3}{68^2}\times\dfrac{34}{95^{\frac{19}{3}}}=\dfrac{(2\cdot 95)^3}{(2\cdot 34)^2}\cdot\dfrac{34}{95^6\cdot 95^{\frac{1}{3}}}=2^{3-2}95^{3-6-\frac{1}{3}}34^{1-2}\\\\=2\cdot 95^{-3\frac{1}{3}}\cdot 34^{-1}=2\cdot 95^{-4+\frac{2}{3}}\cdot 34^{-1}\\\\=\dfrac{2\sqrt[3]{95^2}}{95^4\cdot 34}=\dfrac{\sqrt[3]{95^2}}{17\cdot95^4}\\\\=\dfrac{\sqrt[3]{9\,025}}{1\,384\,660\,625}](https://tex.z-dn.net/?f=%5Cdfrac%7B190%5E3%7D%7B68%5E2%7D%5Ctimes%5Cdfrac%7B34%7D%7B95%5E%7B%5Cfrac%7B19%7D%7B3%7D%7D%7D%3D%5Cdfrac%7B%282%5Ccdot%2095%29%5E3%7D%7B%282%5Ccdot%2034%29%5E2%7D%5Ccdot%5Cdfrac%7B34%7D%7B95%5E6%5Ccdot%2095%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%7D%3D2%5E%7B3-2%7D95%5E%7B3-6-%5Cfrac%7B1%7D%7B3%7D%7D34%5E%7B1-2%7D%5C%5C%5C%5C%3D2%5Ccdot%2095%5E%7B-3%5Cfrac%7B1%7D%7B3%7D%7D%5Ccdot%2034%5E%7B-1%7D%3D2%5Ccdot%2095%5E%7B-4%2B%5Cfrac%7B2%7D%7B3%7D%7D%5Ccdot%2034%5E%7B-1%7D%5C%5C%5C%5C%3D%5Cdfrac%7B2%5Csqrt%5B3%5D%7B95%5E2%7D%7D%7B95%5E4%5Ccdot%2034%7D%3D%5Cdfrac%7B%5Csqrt%5B3%5D%7B95%5E2%7D%7D%7B17%5Ccdot95%5E4%7D%5C%5C%5C%5C%3D%5Cdfrac%7B%5Csqrt%5B3%5D%7B9%5C%2C025%7D%7D%7B1%5C%2C384%5C%2C660%5C%2C625%7D)
Ok so cut the circles into the bottom number! Then which ever is more filled out put a < to it!! Hope this helps tell me if you need more!
