Formula for area:
Area = Wide x Length
Wide: x + 5
Long: 3x - 2
Area = (x + 5) x (3x - 2)
Area = 3x² - 2x + 15x - 10
Area = (3x² + 13x - 10) square feet
I’m pretty sure the answer would be B
Answer:
A. b2 – 4ac = 0
Step-by-step explanation:
Hey user☺☺
Option a is correct
Because the graph has only one solution.
As the graph touches the x-axis at one point that means that it will have only one solution for x. But we know that a quadratic equation has two solutions. So the graph will have two equal solution and therefore the discriminant will be 0.
Hope this will help☺☺
![\bf 343^{\frac{2}{3}}+36^{\frac{1}{2}}-256^{\frac{3}{4}}\qquad \begin{cases} 343=7\cdot 7\cdot 7\\ \qquad 7^3\\ 36=6\cdot 6\\ \qquad 6^2\\ 256=4\cdot 4\cdot 4\cdot 4\\ \qquad 4^4 \end{cases}\\\\\\ (7^3)^{\frac{2}{3}}+(6^2)^{\frac{1}{2}}-(4^4)^{\frac{3}{4}} \\\\\\ \sqrt[3]{(7^3)^2}+\sqrt[2]{(6^2)^1}-\sqrt[4]{(4^4)^3}\implies \sqrt[3]{(7^2)^3}+\sqrt[2]{(6^1)^2}-\sqrt[4]{(4^3)^4} \\\\\\ 7^2+6-4^3\implies 49+6-64\implies -9](https://tex.z-dn.net/?f=%5Cbf%20343%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%2B36%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D-256%5E%7B%5Cfrac%7B3%7D%7B4%7D%7D%5Cqquad%20%5Cbegin%7Bcases%7D%0A343%3D7%5Ccdot%207%5Ccdot%207%5C%5C%0A%5Cqquad%207%5E3%5C%5C%0A36%3D6%5Ccdot%206%5C%5C%0A%5Cqquad%206%5E2%5C%5C%0A256%3D4%5Ccdot%204%5Ccdot%204%5Ccdot%204%5C%5C%0A%5Cqquad%204%5E4%0A%5Cend%7Bcases%7D%5C%5C%5C%5C%5C%5C%20%287%5E3%29%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%2B%286%5E2%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D-%284%5E4%29%5E%7B%5Cfrac%7B3%7D%7B4%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Csqrt%5B3%5D%7B%287%5E3%29%5E2%7D%2B%5Csqrt%5B2%5D%7B%286%5E2%29%5E1%7D-%5Csqrt%5B4%5D%7B%284%5E4%29%5E3%7D%5Cimplies%20%5Csqrt%5B3%5D%7B%287%5E2%29%5E3%7D%2B%5Csqrt%5B2%5D%7B%286%5E1%29%5E2%7D-%5Csqrt%5B4%5D%7B%284%5E3%29%5E4%7D%0A%5C%5C%5C%5C%5C%5C%0A7%5E2%2B6-4%5E3%5Cimplies%2049%2B6-64%5Cimplies%20-9)
to see what you can take out of the radical, you can always do a quick "prime factoring" of the values, that way you can break it in factors to see who is what.
1 out of 52. So 1/52 equals some decimal. Plug in calc.