The first step is to write this equation into general form. The
general form of an equation is:
ax^2 + bx + c = 0
To make this equation to general form, you have to simplify
the equation first.
2/3(x-4) (x+5) = 1
2/3 (x^2 + 5x – 4x – 20) = 1
2/3(x^2 + x -20) = 1
2/3x^2 + 2/3x – 40/3 = 1
2/3x^2 + 2/3x – 40/3 – 1 = 0
2/3x^2 +2/3x – 43/3 = 0
Therefore, a = 2/3 ; b = 2/3 ; c = -43/3
Answer: 24
Step-by-step explanation:
Cube root of 1728 is 12 and on multiplying it by cube root of 14903 we get 295.306.
<u>Solution:
</u>
Need to calculate
and then multiply the result by ![\sqrt[3]{14903}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B14903%7D)
Let us first evaluate ![\sqrt[3]{1728}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B1728%7D)
![\Rightarrow \sqrt[3]{1728}=\sqrt[3]{12 \times 12 \times 12}=12](https://tex.z-dn.net/?f=%5CRightarrow%20%5Csqrt%5B3%5D%7B1728%7D%3D%5Csqrt%5B3%5D%7B12%20%5Ctimes%2012%20%5Ctimes%2012%7D%3D12)
As need to multiply 12 by ![\sqrt[3]{14903}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B14903%7D)
![\Rightarrow 12 \times \sqrt[3]{14903}](https://tex.z-dn.net/?f=%5CRightarrow%2012%20%5Ctimes%20%5Csqrt%5B3%5D%7B14903%7D)
On solving
, we get
![\sqrt[3]{14903}=24.608](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B14903%7D%3D24.608)
![\Rightarrow 12 \times \sqrt[3]{14903}=12 \times 24.608=295.306](https://tex.z-dn.net/?f=%5CRightarrow%2012%20%5Ctimes%20%5Csqrt%5B3%5D%7B14903%7D%3D12%20%5Ctimes%2024.608%3D295.306)
Hence cube root of 1728 is 12 and on multiplying it by cube root of 14903 we get 295.306.
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