Answer:
1) 
2) ![\sqrt[3]{y^5}=y^{\frac{5}{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7By%5E5%7D%3Dy%5E%7B%5Cfrac%7B5%7D%7B3%7D)
3) ![\sqrt[5]{a^{12}}=a^{\frac{12}{5} }](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Ba%5E%7B12%7D%7D%3Da%5E%7B%5Cfrac%7B12%7D%7B5%7D%20%7D)
4) ![\sqrt[4]{z^{9}}=z^\frac{9}{4}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bz%5E%7B9%7D%7D%3Dz%5E%5Cfrac%7B9%7D%7B4%7D)
Step-by-step explanation:
1) 
We know that 
So, 
2) ![\sqrt[3]{y^5}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7By%5E5%7D)
We know that ![\sqrt[3]{x}=x^{\frac{1}{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%7D%3Dx%5E%7B%5Cfrac%7B1%7D%7B3%7D)
So, ![\sqrt[3]{y^5}=y^{\frac{5}{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7By%5E5%7D%3Dy%5E%7B%5Cfrac%7B5%7D%7B3%7D)
3) ![\sqrt[5]{a^{12}}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Ba%5E%7B12%7D%7D)
We know that ![\sqrt[5]{x}=x^{\frac{1}{5}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Bx%7D%3Dx%5E%7B%5Cfrac%7B1%7D%7B5%7D)
So, ![\sqrt[5]{a^{12}}=a^{\frac{12}{5} }](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Ba%5E%7B12%7D%7D%3Da%5E%7B%5Cfrac%7B12%7D%7B5%7D%20%7D)
4) ![\sqrt[4]{z^{9}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bz%5E%7B9%7D%7D)
We know that ![\sqrt[4]{x}=x^{\frac{1}{4}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%7D%3Dx%5E%7B%5Cfrac%7B1%7D%7B4%7D)
So, ![\sqrt[4]{z^{9}}=z^\frac{9}{4}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bz%5E%7B9%7D%7D%3Dz%5E%5Cfrac%7B9%7D%7B4%7D)
Answer:
2b(a+b)
Step-by-step explanation:
(a+b)²-a²+b²=a²+2ab+b²-a²+b²=2ab+2b²=2b(a+b)
So, an isosceles triangle has 2 equal sides, we will call them x.
The base can be modeled by 3+x
So, the perimeter is equal to x + 3+x + x
So
15.6 = 3x + 3
12.6 = 3x
4.2 = x
So, the lengths of the sides are 4.2, 4.2, 7.2
(x+3)(x-1/4)=x^2+11/4-3/4
The equation of a circle is (x-h)²+(y-k)²=r², with h being the x value in the center, k being the y value, and r being the radius. x and y can stay variables, so we can plug numbers in to get (x-0)²+(y-3)²=5²=25=x²+(y-3)²