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Olin [163]
2 years ago
5

Triangle TUV is dilated by a scale factor of 5 to form triangle T'U'V'. Side VT

Mathematics
2 answers:
worty [1.4K]2 years ago
8 0

Answer: 65

Step-by-step explanation: VT•5= v’T’

13•5= V’T’ 13•5=65

Misha Larkins [42]2 years ago
7 0

Answer:

Step-by-step explanationinternet can be used

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I believe he slept for 9 hours of the day

Step-by-step explanation:

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2 years ago
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IgorC [24]

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b

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3 years ago
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3 years ago
Find an n^th degree polynomial with real coefficients satisfying the given conditions. n = 3; -2 and 2 i are zeros; f(-1) = 15.
Ira Lisetskai [31]
So, n = 3, is a 3rd degree polynomial, roots are -2 and 2i

well 2i is a complex root, or imaginary, and complex root never come all by their lonesome, their sister is always with them, the conjugate, so if 0+2i is there, 0-2i is there too

so, the roots are -2, 2i, -2i

now... \bf \begin{cases}&#10;x=-2\implies x+2=0\implies &(x+2)=0\\&#10;x=2i\implies x-2i=0\implies &(x-2i)=0\\&#10;x=-2i\implies x+2i=0\implies &(x+2i)=0&#10;\end{cases}&#10;\\\\\\&#10;(x+2)\underline{(x-2i)(x+2i)}=0\\\\&#10;-----------------------------\\\\&#10;\textit{difference of squares}&#10;\\ \quad \\&#10;(a-b)(a+b) = a^2-b^2\qquad \qquad &#10;a^2-b^2 = (a-b)(a+b)\\\\&#10;-----------------------------\\\\&#10;(x+2)[x^2-(2i)^2]=0\implies (x+2)[x^2-(2^2i^2)]=0&#10;\\\\\\&#10;(x+2)[x^2-(4\cdot -1)]=0\implies (x+2)(x^2+4)=0&#10;\\\\\\&#10;x^3+2x^2+4x+8=0

now, if we check f(-1), we end up with 5, not 15
hmmm

so, how to turn our 5 to 15? well, 3*5, thus

\bf 3(x^3+2x^2+4x+8)=f(x)\implies 3(5)=f(-1)\implies 15=f(-1)

usually, when we get the roots, or zeros, if any common factor that is a constant is about, they get in a division with 0 and get tossed, and aren't part of the roots, thus, we can simply add one, in this case, the common factor of 3, to make the 5 turn to 15
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3 years ago
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Gennadij [26K]

Answer:

Step-by-step explanation:

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