The first option is correct. For the following question, the simplified expression for the volume is 8x3 + 14x2 + 3x.
Answer:
The range of the graph is [-10, 10]
Step-by-step explanation:
This represents how the y-values of the graph range from -10 to 10.
Because there are no arrow marks on the ends of the graph, we know that the graph stops there at -10 and 10 and doesn't go beyond that.
Since AB = CD the trapezoid is isosceles, which means that ∡A = ∡D
Therefore also ∡2 = ∡3 (they are half of the congruent angles)
For the properties of parallel lines (BD and AD) crossed by a transversal (BD) we have ∡3 = ∡CBD.
Now consider triangles AOD and BCD:
∡OAD (2) = ∡ADO (3) = ∡CBD (3) = ∡CDB (4)
T<span>he sum of the angles of a triangle must be 180°, t</span>herefore:
∡AOD = 180 - ∡2 - ∡3
∡BCD = 180 - ∡3 - ∡4
∡AOD = ∡BCD because their measure is the difference of congruent angles.
The normal vector to the plane <em>x</em> + 3<em>y</em> + <em>z</em> = 5 is <em>n</em> = (1, 3, 1). The line we want is parallel to this normal vector.
Scale this normal vector by any real number <em>t</em> to get the equation of the line through the point (1, 3, 1) and the origin, then translate it by the vector (1, 0, 6) to get the equation of the line we want:
(1, 0, 6) + (1, 3, 1)<em>t</em> = (1 + <em>t</em>, 3<em>t</em>, 6 + <em>t</em>)
This is the vector equation; getting the parametric form is just a matter of delineating
<em>x</em>(<em>t</em>) = 1 + <em>t</em>
<em>y</em>(<em>t</em>) = 3<em>t</em>
<em>z</em>(<em>t</em>) = 6 + <em>t</em>
