Which of these strategies would eliminate a variable in the system of equations?
1 answer:
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Answer:
Area = 13.15 square units
Step-by-step explanation:
Let the given vertices be represented as follows:
A(2, -1, 1) = 2i - j + k
B(5, 1, 4) = 5i + j + 4k
C(0, 1, 1) = 0i + j + k
D(3, 3, 4) = 3i + 3j + 4k
(i) Let's calculate the vectors of all the sides:
AB = B - A = (5i + j + 4k) - (2i - j + k)
AB = 5i + j + 4k - 2i + j - k [Collect like terms]
AB = 3i + 2j + 3k
BC = C - B = (0i + j + k) - (5i + j + 4k)
BC = 0i + j + k - 5i - j - 4k [Collect like terms]
BC = -5i + 0j - 3k
CD = D - C = (3i + 3j + 4k) - (0i + j + k)
CD = 3i + 3j + 4k - 0i - j - k [Collect like terms]
CD = 3i + 2j + 3k
DA = A - D = (2i - j + k) - (3i + 3j + 4k)
DA = 2i - j + k - 3i - 3j - 4k [Collect like terms]
DA = -i - 4j - 3k
AC = C - A = (0i + j + k) - (2i - j + k)
AC = 0i + j + k - 2i + j - k [Collect like terms]
AC = -2i + 2j
BD = D - B = (3i + 3j + 4k) - (5i + j + 4k)
BD = 3i + 3j + 4k - 5i - j - 4k [Collect like terms]
BD = -2i + 2j
(ii) From the results in (i) above, we can deduce that;
AB = CD This implies that AB || CD [AB is parallel to CD]
AC = BD This implies that AC || BD [AC is parallel to BD]
(iii) Therefore, ABDC is a parallelogram since opposite sides (AB and CD) are parallel. Hence, the points are vertices of a parallelogram
<u>Now let's calculate the area</u>
To find the area of the parallelogram, we find the magnitude of the cross product of any two adjacent sides.
In this case, we'll choose AB and AC
Area = |AB X AC|
Where;
<u></u>
AB X AC = i(0 - 6) - j(0 + 6) + k(6 + 4)
AB X AC = - 6i - 6j + 10k
|AB X AC| =
|AB X AC| =
|AB X AC| = 13.15
Area = 13.15 square units.
<u></u>
<u></u>
<u>PS: </u> ACBD is also a parallelogram. The diagram has also been attached to this response.
Answer:
Step-by-step explanation:
<u>Part A: the x-intercepts are -3/2 and 5/2 </u>
- 4
- 7x - 15 -----> (2x + 3) (2x - 5)
2x + 3 = 0 --------> 2x = -3 ----------> x =
2x - 5 = 0 ---------> 2x = 5 -----------> x =
<u>Part B: the parabola is a minimum, the vertex is (7/8,-289/16) </u>
f(x)=ax^2+bx+c
if a>0, then the parabola opens up and the vertex is a minimum
a<0 then the parabola opens down and the vertex is a max
- a = 4, it is a minimum
the x value of the vertex in f(x)=ax^2+bx+c= is -b/(2a)
the y value of the vertex is f(-b/(2a))
f(x) = 4 - 7x - 15
a = 4
b = -7
-b/2a=-(7)/(2*4)=7/8
f(7/8) = 4 - 7(7/8) - 15
f(7/8) = 49/16 - 49/8 - 15
f(7/8) = -289/16
- the vertex is (7/8,-289/16)
<u>Part C: </u>
- the vertex is minimum and the graph goes through the x intercepts
- plug in x values to get the y value (EX: choose 0 for x, and you'll get -15 for y, so you would plot the point at (0, -15). you could plug in 1 for x, and get -18 for y, plotting the point (1, -18) )