I hope the choices for the numerators of the solutions are given.
I am showing the complete work to find the solutions of this equation , it will help you to find an answer of your question based on this solution.
The standard form of a quadratic equation is :
ax² + bx + c = 0
And the quadratic formula is:
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
So, first step is to compare the given equation with the above equation to get the value of a, b and c.
So, a = 10, b = -19 and c = 6.
Next step is to plug in these values in the above formula. Therefore,
So,
So,
Hope this helps you!
Answer:
(x − 2) = 3
Remove the bracket
x - 2 = 3
Group the constants at one side of the equation
x = 2 + 3
x = 5
Therefore x = 5.00 to 3 significant figures.
Hope this helps
let the price of pizza and soda( per 2 litre bottle) be x and y
from 1st condition
12x+2y=137.6.....(1)
from 2nd condition
x+y =11.95.......(2)
multiply eq. (2) by 2
2x+2y=23.9.....(3)
subtracting (1) and (3)
12x+2y=137.6
2x+2y= 23.9
10x = 113.7
x=113.7/10
x=11.37
2(11.37)+2y=23.9
22.74+2y=23.9
2y=23.9-22.7
2y=1.2
y=0.6
So the prices of the pizza and soda are 11.37 and 0.6 dollars.
You might need a maths degree for this one, I believe it’s 4 but don’t quote me on that.
Answer: ![3x^2y\sqrt[3]{y}\\\\](https://tex.z-dn.net/?f=3x%5E2y%5Csqrt%5B3%5D%7By%7D%5C%5C%5C%5C)
Work Shown:
![\sqrt[3]{27x^{6}y^{4}}\\\\\sqrt[3]{3^3x^{3+3}y^{3+1}}\\\\\sqrt[3]{3^3x^{3}*x^{3}*y^{3}*y^{1}}\\\\\sqrt[3]{3^3x^{2*3}*y^{3}*y}\\\\\sqrt[3]{\left(3x^2y\right)^3*y}\\\\\sqrt[3]{\left(3x^2y\right)^3}*\sqrt[3]{y}\\\\3x^2y\sqrt[3]{y}\\\\](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B27x%5E%7B6%7Dy%5E%7B4%7D%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B3%5E3x%5E%7B3%2B3%7Dy%5E%7B3%2B1%7D%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B3%5E3x%5E%7B3%7D%2Ax%5E%7B3%7D%2Ay%5E%7B3%7D%2Ay%5E%7B1%7D%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B3%5E3x%5E%7B2%2A3%7D%2Ay%5E%7B3%7D%2Ay%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B%5Cleft%283x%5E2y%5Cright%29%5E3%2Ay%7D%5C%5C%5C%5C%5Csqrt%5B3%5D%7B%5Cleft%283x%5E2y%5Cright%29%5E3%7D%2A%5Csqrt%5B3%5D%7By%7D%5C%5C%5C%5C3x%5E2y%5Csqrt%5B3%5D%7By%7D%5C%5C%5C%5C)
Explanation:
As the steps above show, the goal is to factor the expression under the root in terms of pulling out cubed terms. That way when we apply the cube root to them, the exponents cancel. We cannot factor the y term completely, so we have a bit of leftovers.
