Factor 4x^4yz-16y^3z

Mathematics

1 answer:

Iteru [2.4K]3 years ago

6 0

Answer:

Step-by-step explanation:

$4x^4yz-16y^3z\\\\4x^4yz=4yz\cdot x^4\\\\16y^3z=4yz\cdot4y^2\\\\4x^4yz-16y^3z=4yz\cdot x^4-4yz\cdot4y^2=4yz(x^4-4y^2)\\\\x^4-4y^2=x^{2\cdot2}-2^2y^2=(x^2)^2-(2y)^2=(x^2-2y)(x^2+2y)\\\\Used:\\\\(a^n)^m=a^{nm}\\\\(ab)^n=a^nb^n\\\\a^2-b^2=(a-b)(a+b)\\\\4x^4yz-16y^3z=4yz(x^2-2y)(x^2+2y)$

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Step-by-step explanation:

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kolbaska11 [484]

Answer:

$\csc \left(\theta-\frac{\pi }{2}\right)=0.73$

Step-by-step explanation:

The identity you will use is:

$\csc \left(x\right)=\frac{1}{\sin \left(x\right)}$

So,

$\csc \left(\theta-\frac{\pi }{2}\right)$

$\csc \left(\theta-\frac{\pi }{2}\right)=\frac{1}{\sin \left(-\frac{\pi }{2}+\theta\right)}$

Now, using the difference of sin

Note: state that $\text{sin}(\alpha\pm \beta)=\text{sin}(\alpha) \text{cos}(\beta) \pm \text{cos}(\alpha) \text{sin}(\beta)$

$\csc \left(\theta-\frac{\pi }{2}\right)=\frac{1}{-\cos \left(\theta\right)\sin \left(\frac{\pi }{2}\right)+\cos \left(\frac{\pi }{2}\right)\sin \left(\theta\right)}$