Which of the following is the correct factorization of the polynomial below?

Mathematics

2 answers:

Archy [21]3 years ago

8 0

the answer is D- unreduceable.

vitfil [10]3 years ago

7 0

x^3 - 18

This polynomial is not factorable

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Use counters to find the quotient and remainder for 36 divide a by 8

koban [17]

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Ket [755]

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Step-by-step explanation:

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Find the volume of the solid bounded by z = 2 - x2 - y2 and z = 1. Express your answer as a decimal rounded to the hundredths pl

butalik [34]

Answer:

The answer is " $(\frac{\pi}{2})$ ".

Step-by-step explanation:

$z = 2 - x^2 - y^2.........(1) \\\\z = 1.............(2)$

Let add equation 1 and 2:

Using formula: $x^2+y^2=1 \\\\$

convert to polar coordinates

$r=2$

$\theta \varepsilon (0=\pi)=z\\\\V=\int^{2\pi}_{\theta=0}\int^{1}_{\pi=0}\int^{z_{2}}_{z_1} r \ dr \d \theta\\\\$

$=\int^{2\pi}_{0}\int^{1}_{0} (Z_2-z_1)r \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} 1- (-1-x^2-y^2) r \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} (+1 \pm r^2) r \ dr \d \theta\\\\=\int^{2\pi}_{0}\int^{1}_{0} (-r^3 + r) \ dr \d \theta\\\\=\int^{2\pi}_{0} (-\frac{r^4}{4}+\frac{r^2}{1})^{1}_{0} \d \theta\\\\=\int^{2\pi}_{0} (\frac{1}{4}) \d \theta\\\\=(\frac{2 \pi}{4}) \\\\=(\frac{\pi}{2}) \\\\$