-2x + 16 < = 10
-2x < = 10 - 16
-2x < = - 6
x > = -6/-2
x > = 3............so any number that is 3 or above is in the solution set
-8.1
Step-by-step explanation:

<em>Times </em><em>all </em><em>by </em><em>9</em><em> </em><em>to </em><em>get </em><em>rid </em><em>of </em><em>fraction</em>
<em>
</em>
<em>Take </em><em>2</em><em>1</em><em>.</em><em>6</em><em> </em><em>away</em><em> </em><em>from</em><em> </em><em>both</em><em> </em><em>sides</em>
<em>
</em>
<em>Divide</em><em> </em><em>by </em><em>-</em><em>4</em>
<em>
</em>
Answer:
Remember if cosecθ=1/sinθ
sinθ×cosecθ=sinθ×1/sinθ=1
CMIIW
Answer:
- zeros: x = -3, -1, +2.
- end behavior: as x approaches -∞, f(x) approaches -∞.
Step-by-step explanation:
I like to use a graphing calculator for finding the zeros of higher order polynomials. The attachment shows them to be at x = -3, -1, +2.
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The zeros can also be found by trial and error, trying the choices offered by the rational root theorem: ±1, ±2, ±3, ±6. It is easiest to try ±1. Doing so shows that -1 is a root, and the residual quadratic is ...
x² +x -6
which factors as (x -2)(x +3), so telling you the remaining roots are -3 and +2.
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For any odd-degree polynomial with a positive leading coefficient, the sign of the function will match the sign of x when the magnitude of x gets large. Thus as x approaches negative infinity, so does f(x).