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Anuta_ua [19.1K]
3 years ago
6

Writethecoordinatesoftheverticesafteradilationwithascalefactorof 1 5 ,centeredattheorigin. y x -10 10 -10 10 0 S T U S(5, -10) →

S′( , ) T(10, -10) → T′( , ) U(5, 10) → U′( , )
Mathematics
2 answers:
densk [106]3 years ago
6 0

Answer:

The answer is below

Step-by-step explanation:

Write the coordinates of the vertices after a dilation with a scale factor of 1/5 , centered at the origin.

Transformation is the movement of a point from its initial location to a new location. Types of transformation are rotation, reflection, dilation and translation.

Dilation is the reduction or enlargement in the size of an object by a scale factor (k). If k > 1, it is an enlargement and if k < 1, it is a reduction. If a point A(x, y) is dilated by a factor k, the new point is A'(kx, ky).

Therefore, if the vertices are dilated with a scale factor of 1/5 , centered at the origin. The new point is:

S(5, -10) → S′(1 , -2) T(10, -10) → T′(2 , -2) U(5, 10) → U′(1 , 2)

riadik2000 [5.3K]3 years ago
4 0

Answer:

j

j

Step-by-step explanation:

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Feliz [49]

Recall Euler's theorem: if \gcd(a,n) = 1, then

a^{\phi(n)} \equiv 1 \pmod n

where \phi is Euler's totient function.

We have \gcd(9,32) = 1 - in fact, \gcd(9,32^k)=1 for any k\in\Bbb N since 9=3^2 and 32=2^5 share no common divisors - as well as \phi(9) = 6.

Now,

37^{32} = (1 + 36)^{32} \\\\ ~~~~~~~~ = 1 + 36c_1 + 36^2c_2 + 36^3c_3+\cdots+36^{32}c_{32} \\\\ ~~~~~~~~ = 1 + 6 \left(6c_1 + 6^3c_2 + 6^5c_3 + \cdots + 6^{63}c_{32}\right) \\\\ \implies 32^{37^{32}} = 32^{1 + 6(\cdots)} =  32\cdot\left(32^{(\cdots)}\right)^6

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32^{37^{32}} \equiv 32\cdot1 \equiv \boxed{5} \pmod9

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Step-by-step explanation:

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