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yan [13]
3 years ago
7

One of the factors of x² - x – 6 is How do i work this out on paper?

Mathematics
2 answers:
rewona [7]3 years ago
4 0
X^2 -x- 6 equals (x-3)(x+2). Just look at the negative six and think, what two numbers give me this? Then, get a pair of what you find, and see if the two numbers added equal to negative one (which comes from the negative "x" as the coefficient is always 1). That's how you get -3 and +2.
Ulleksa [173]3 years ago
3 0
Could you re-write that for me?
I'm not sure I'm understanding it right.

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Find the volume of the solid under the plane 5x + 9y − z = 0 and above the region bounded by y = x and y = x4.
svp [43]
<span>For the plane, we have z = 5x + 9y

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The volume of the solid equals

\int\limits^1_0 { \int\limits_{x^4}^x {(5x+9y)} \, dy } \, dx = \int\limits^1_0 {\left[5xy+ \frac{9}{2} y^2\right]_{x^4}^{x}} \, dx  \\  \\ =\int\limits^1_0 {\left[\left(5x(x)+ \frac{9}{2} (x)^2\right)-\left(5x(x^4)+ \frac{9}{2} (x^4)^2\right)\right]} \, dx  \\  \\ =\int\limits^1_0 {\left(5x^2+ \frac{9}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx =\int\limits^1_0 {\left( \frac{19}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx \\  \\ =\left[ \frac{19}{6} x^3- \frac{5}{6} x^6- \frac{1}{2} x^9\right]^1_0

=\frac{19}{6} - \frac{5}{6} - \frac{1}{2} =\bold{ \frac{11}{6} \ cubic \ units}</span>
8 0
3 years ago
Hard time with these type of problems, need help
amid [387]
Not sure I have had idea of what it is but I don’t wanna give u the wrong answer
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