If the equation is (x + 1)(2x − 3) will the solution have two different lengths.
1 answer:
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From the right hand side, we will need to find a way to rewriting 3x²y in terms of cube roots.
We know that 27 is 3³, so if we were to rewrite it in terms of cube roots, we will need to multiply everything by itself two more twice. (ie we can rewrite it as ∛(3x²y)³)
Hence, we can say that it's:
![\sqrt[3]{162x^{c}y^{5}} = \sqrt[3]{(3x^{2}y)^{3}} * \sqrt[3]{6y^{d}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B162x%5E%7Bc%7Dy%5E%7B5%7D%7D%20%3D%20%5Csqrt%5B3%5D%7B%283x%5E%7B2%7Dy%29%5E%7B3%7D%7D%20%2A%20%5Csqrt%5B3%5D%7B6y%5E%7Bd%7D%7D)
![= \sqrt[3]{162x^{6}y^{3+d}}](https://tex.z-dn.net/?f=%3D%20%5Csqrt%5B3%5D%7B162x%5E%7B6%7Dy%5E%7B3%2Bd%7D%7D)
Hence, c = 6 and d = 2
We know that 27 is 3³, so if we were to rewrite it in terms of cube roots, we will need to multiply everything by itself two more twice. (ie we can rewrite it as ∛(3x²y)³)
Hence, we can say that it's:
Hence, c = 6 and d = 2
Answer:
4x +y = 3
Step-by-step explanation:
Perpendicular lines have slopes that are the negative reciprocals of one another. When the equation of the line is written in standard form like this, the equation of the perpendicular line can be written by swapping the x- and y-coefficients and negating one of them. Doing this much would give you ...
4x +y = (constant)
Note that we have chosen to make the equation read 4x+y, not -4x-y. The reason is that "standard form" requires the leading coefficient to be positive.
Now, you just need to make sure the constant is appropriate for the point you want the line to go through. So, it needs to be ...
4(2) +(-5) = constant = 3
The line of interest has equation ...
4x + y = 3
