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Law Incorporation [45]
3 years ago
8

A triangle has vertices at B(-3,0), C(2, -1), D(-1,2). Which series of transformations would produce an image with vertices B"(4

,1), C"(-1,0), D"(2, 3)?
O(x, y) - (X. -y). (x, y) (x + 1. y + 1)
O(x, y) - (-x, y). (X. y) (x + 1, y + 1)
O(x, y) - (x, -y), (x, y) (X + 2, Y + 2)
O(x, y) - (-x, y), (x, y) + (x + 2, Y + 2)
Mathematics
1 answer:
Alik [6]3 years ago
4 0

Answer:

  (b)  (x, y) ⇒ (-x, y); (x, y) ⇒ (x + 1, y + 1)

Step-by-step explanation:

A graph shows the image is consistent with reflection over the y-axis

  (x, y) ⇒ (-x, y)

and translation right 1, up 1

  (x, y) ⇒ (x +1, y +1)

These transformations are listed in the second choice.

__

In the attachment, the original image is blue, the reflected image is purple, and the final translated image is red.

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Step-by-step explanation:

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2 years ago
Use the fraction strips to compare 2/4 and 5/8. use the drop-down menus to explain your comparison.
tia_tia [17]

Answer:

2

5

\dfrac{4}{5}

Step-by-step explanation:

We are given 2 fractional numbers:

1.\ \dfrac{2}4\\2.\ \dfrac{5}8

We have to use fraction strips to compare to the fractional numbers.

Let we are Comparing \frac{2}{4} with the length of x number of \frac{1}4 sections.

i.e.

\dfrac{2}{4}  = x \times \dfrac{1}{4}\\\Rightarrow x = \dfrac{2 \times 4}{4}\\\Rightarrow x = 2

Let we are Comparing \frac{5}{8} with the length of y number of \frac{1}8 sections.

i.e.

\dfrac{5}{8}  = y \times \dfrac{1}{8}\\\Rightarrow y = \dfrac{5 \times 8}{8}\\\Rightarrow y = 5

Now, let us have a look at 3rd part of question:

The sections of 2/4 is _____ the length of 5/8. Therefore, 2/4 < 5/8

Let the answer be z.

So, the equation becomes:

\dfrac{2}{4} = z \times \dfrac{5}{8}\\\Rightarrow z = \dfrac{2 \times 8}{4 \times 5}\\\Rightarrow z = \dfrac{2 \times 2}{5}\\\Rightarrow z = \dfrac{4}{5}

So, the answers are:

2

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\dfrac{4}{5}

5 0
3 years ago
What is the volume of the composite figure shown? (Use 3.14 for π.)
Nimfa-mama [501]
The\ volume\ of\ the\ cone:V_1=\frac{1}{3}\pi r^2 H\\(r-a\ radius;\ H-height)\\------------------\\r=7ft;\ H=7ft\\\\V_1=\frac{1}{3}\pi\cdot7^2\cdot7=\frac{1}{3}\pi\cdot343=\frac{343}{3}\pi\ (ft^2)\\-------------------\\The\ volume\ of\ the\ cylinder:V_2=\pi r^2 H\\(r-a\ radius;\ H-height)\\-------------------\\r=7ft;\ H=6ft\\\\V_2=\pi\cdot7^2\cdot6=294\pi\ (ft^3)\\------------------------\\The\ volume\ of\ the\ composite\ figure: V_F=V_1+V_2

V_F=\frac{343}{3}\pi+294\pi=\frac{343}{3}\pi+\frac{294\cdot3}{3}\pi=\frac{343}{3}\pi+\frac{882}{3}\pi=\boxed{\frac{1225}{3}\pi\ (ft^3)}\\\\\approx\frac{1225}{3}\cdot3.14}=\frac{3846.5}{3}\approx\boxed{1,282.17\ (ft^3)}\leftarrow answer
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