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2 years ago

Which of the following steps were applied to ABCD to obtain A’B’C’D’?

Mathematics

2 answers:

devlian [24]2 years ago

6 0

Answer:

c- shifted 3 units right and 4 units up

Step-by-step explanation:

In this problem, we have a quadrilateral named as ABCD. Recall that a quadrilateral is a two-dimensional shape having four sides. So, we need to identify what transformation has been performed to get A'B'C'D', which is the same quadrilateral shifted certain units right and up. So take one point, say, B, so how do we need to do to obtain point B'? well, we need to move that point 3 units right and 4 units up, but how can we know this? just count the number of squares you need to move from B to B' horizontally and vertically, which is in fact 3 units right and 4 units up.

Murrr4er [49]2 years ago

4 0

Answer: The correct option is

(c) shifted 3 units right and 4 units up

Step-by-step explanation: We are given to select the correct steps that were applied o ABCD to obtain A'B'C'D' as shown in the figure.

From the graph, we note that

the co-ordinates of the vertices of quadrilateral ABCD are A(2, 3), B(4, 8), C(6, 8) and D(8, 3).

And, the co-ordinates of the vertices of quadrilateral A'B'C'D' are A'(5, 7), B'(7, 12), C'(9, 12) and D'(11, 7).

So, the transformations from the vertices of ABCD to the vertices of A'B'C'D' are as follows :

A(2, 3) ⇒ A'(5, 7) = (2+3, 3+4),

B(4, 8) ⇒ B'(7, 12) = (4+3, 8+4),

C(6, 8) ⇒ C'(9, 12) = (6+3, 8+4),

D(8, 3) ⇒ D'(11, 7) = (8+3, 3+4).

Therefore, the required rule of translation from ABCD to A'B'C'D' is

(x, y) ⇒ (x+3, y+4). That is, 3 units right and 4 units up.

Thus, the required steps applied to ABCD to obtain A'B'C'D' are

Shifted 3 units right and 4 units up.

Option (c) is CORRECT.

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Two forces, F1 and F2, are represented by vectors with initial points that are at the origin. The first force has a magnitude of

Strike441 [17]

Answer:

Step-by-step explanation:

The two force F1 and F2 are represented by vectors with initial points that are at the origin.

the terminal point of the vector is point P(1, 1, 0)

Therefore, the direction of the vector force is

$v_1=(1-0)\hat i+(1-0)\hat j +(0-0)\hat k\\\\=1\hat i+1\hat j+0\hat k\\\\=\hat i + \hat j$

The unit vector in the direction of force will be

$\frac{v_1}{|v_1|} =\frac{\hat i+ \hat j}{\sqrt{1^2+1^2+0^2} } \\\\=\frac{1}{\sqrt{2} (\hat i +\hat j)}$

The magnitude of the force is 40lb, so the force will be

$F_1=40\times \frac{1}{\sqrt{2} } (\hat i+\hat j)\\\\=20\sqrt{2} (\hat i+\hat j)$

The terminal point of its vector is point Q(0, 1, 1)

Therefore, the direction of the vector force is

$v_2=(0-0)\hat i+(1-0)\hat j +(1-0)\hat k\\\\=0\hat i+1\hat j+1\hat k\\\\=\hat j + \hat k$

$\frac{v_2}{|v_2|} =\frac{\hat j+ \hat k}{\sqrt{0^2+1^2+1^2} } \\\\=\frac{1}{\sqrt{2} (\hat j +\hat k)}$

The magnitude of the force is 60lb, so the force will be

$F_2=60\times \frac{1}{\sqrt{2} } (\hat j+\hat k)\\\\=30\sqrt{2} (\hat j+\hat k)$

The resultant of the two forces is

$F=F_1+F_2\\\\=[20\sqrt{2} (\hat i+\hat j)]+[30\sqrt{2} (\hat j +\hat k)]\\\\=20\sqrt{2} \hat i+20\sqrt{2} \hat j +30\sqrt{2} \hat j+30\sqrt{2} \hat k\\\\=20\sqrt{2} \hat i+50\sqrt{2} \hat j+30\sqrt{2} \hat k$