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Romashka-Z-Leto [24]
3 years ago
13

What sequence of transformations, when applied to △XYZ , shows that △XYZ is similar to △X′Y′Z′ ?

Mathematics
2 answers:
jek_recluse [69]3 years ago
8 0

Answer:

dilation with respect to the origin by a scale factor of 2 followed by a translation of 4 units down

Step-by-step explanation:

stiks02 [169]3 years ago
4 0
The coordinates of Δxyz ⇒⇒ x(1,1)     y(2,3)    z(3,2)
The coordinates of Δx'y'z' ⇒⇒ x'(2,-2)   y'(4,2)    z'(6,0)

1. Applying the transformation of the first sentence on Δxyz
The results are ⇒⇒⇒ (2,-2)   (4,2)    (6,0)

2. Applying the transformation of the second sentence on Δxyz
The results are ⇒⇒⇒ (12,14)   (24,38)    (36,28)

3. Applying the transformation of the third sentence on Δxyz
The results are ⇒⇒⇒ (12,-36)   (24,-12)    (36,-24)

4. Applying the transformation of the fourth sentence on Δxyz
The results are ⇒⇒⇒ (2,-6)   (4,-2)    (6,-4)

By comparing the results obtained with the coordinates of Δx'y'z'
So, the correct answer is the first sentence.
<span>dilation with respect to the origin by a scale factor of 2 followed by a translation of 4 units down</span>








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What is the interest earned based upon a principal of $1,000 at a rate of 6% over 8 years????
Nata [24]
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A changes saved<br> Set up a proportion and use it to solve for x.<br> 2<br> 10<br> х<br> 3
V125BC [204]

Answer:

C. 15

Step-by-step explanation:

The following proportional can be set-up for the figure given:

\frac{x + 3}{x} = \frac{10 + 2}{10}

Solve for x

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3 0
3 years ago
Consider the following set of vectors. v1 = 0 0 0 1 , v2 = 0 0 3 1 , v3 = 0 4 3 1 , v4 = 8 4 3 1 Let v1, v2, v3, and v4 be (colu
Alona [7]

Answer:

We have the equation

c_1\left[\begin{array}{c}0\\0\\0\\1\end{array}\right] +c_2\left[\begin{array}{c}0\\0\\3\\1\end{array}\right] +c_3\left[\begin{array}{c}0\\4\\3\\1\end{array}\right] +c_4\left[\begin{array}{c}8\\4\\3\\1\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]

Then, the augmented matrix of the system is

\left[\begin{array}{cccc}0&0&0&8\\0&0&4&4\\0&3&3&3\\1&1&1&1\end{array}\right]

We exchange rows 1 and 4 and rows 2 and 3 and obtain the matrix:

\left[\begin{array}{cccc}1&1&1&1\\0&3&3&3\\0&0&4&4\\0&0&0&8\end{array}\right]

This matrix is in echelon form. Then, now we apply backward substitution:

1.

8c_4=0\\c_4=0

2.

4c_3+4c_4=0\\4c_3+4*0=0\\c_3=0

3.

3c_2+3c_3+3c_4=0\\3c_2+3*0+3*0=0\\c_2=0

4.

c_1+c_2+c_3+c_4=0\\c_1+0+0+0=0\\c_1=0

Then the system has unique solution that is (c_1,c_2c_3,c_4)=(0,0,0,0) and this imply that the vectors v_1,v_2,v_3,v_4 are linear independent.

4 0
3 years ago
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