Greeting!
15x² = x + 2
15x² - x -2= 0
Factor left side
(5x-2)(3x+1)=0
Set factors equal to 0
5x - 2 = 0 or 3x + 1 -0
x = 2/5 or x= -1/3
I hope that helps!
Brainlest answer :)
Answer:
![\sqrt[3]{x^{5}y} = x^{\frac{5}{3}}y^\frac{1}{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E%7B5%7Dy%7D%20%3D%20x%5E%7B%5Cfrac%7B5%7D%7B3%7D%7Dy%5E%5Cfrac%7B1%7D%7B3%7D)
Step-by-step explanation:
Given
![\sqrt[3]{x^{5}y}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E%7B5%7Dy%7D)
Required
The equivalent expression
We have:
![\sqrt[3]{x^{5}y}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E%7B5%7Dy%7D)
Rewrite as an exponent
![\sqrt[3]{x^{5}y} = (x^{5}y)^\frac{1}{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E%7B5%7Dy%7D%20%3D%20%28x%5E%7B5%7Dy%29%5E%5Cfrac%7B1%7D%7B3%7D)
Open bracket
![\sqrt[3]{x^{5}y} = x^{5*\frac{1}{3}}y^\frac{1}{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E%7B5%7Dy%7D%20%3D%20x%5E%7B5%2A%5Cfrac%7B1%7D%7B3%7D%7Dy%5E%5Cfrac%7B1%7D%7B3%7D)
![\sqrt[3]{x^{5}y} = x^{\frac{5}{3}}y^\frac{1}{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E%7B5%7Dy%7D%20%3D%20x%5E%7B%5Cfrac%7B5%7D%7B3%7D%7Dy%5E%5Cfrac%7B1%7D%7B3%7D)
Answer:
Step-by-step explanation:
A. The perimeter of quadrilaterals are same when x=2.
B. The area of quadrilaterals are same when x=4.6
<u>Step-by-step explanation</u>:
A. <u>Perimeter of the quadrilaterals must be equal.</u>
- Perimeter of square is 4a.
- where, a = 6
- Perimeter of rectangle is 2(l+b).
- where, l = 2 and b = (3x+4)
Perimeter of square = Perimeter of rectangle
4(6) = 2(2+3x+4)
⇒ 24 = 2(3x+6)
⇒ 24 = 6x + 12
⇒ 6x = 12
⇒ x = 12/6
x = 2
For x=2, the perimeters of both the quadrilaterals are same.
B. <u>Area of the quadrilaterals must be equal.</u>
- Area of square is a².
- where, a = 6.
- Area of rectangle is l
b. - where, l = 2 and b = (3x+4)
Area of square = Area of rectangle
(6)² = 2(3x+4)
36 = 6x+8
6x = 28
x = 28/6
x = 4.6
For x=4.6, the area of both the quadrilaterals are same.