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Verdich [7]
3 years ago
8

Let f(x)=2x^2-x-1 and let g(x)=x-1 Which statement is true?​

Mathematics
1 answer:
sergij07 [2.7K]3 years ago
7 0

Answer:

B is true

Step-by-step explanation:

given f(x) = 2x² - x - 1 and g(x) = x - 1

A

If x = - 1 is a root then f(- 1) = 0

f(- 1) = 2(- 1)² - (- 1) - 1 = 2 + 1 - 1 = 2 ≠ 0 ← False

B

If (x - 1) is a factor then x = 1 is a root and f(1) = 0

f(1) = 2(1)² - 1 - 1 = 2 - 1 - 1 = 0

⇒ g(x) = x - 1 is a factor of f(x) ← True

C

If x = 2 is a root then f(2) = 0

f(2) = 2(2)² - 2 - 1 = 8 - 3 = 5 ≠ 0 ← False

D

\frac{x2x^2-x-1}{x-1} = \frac{(2x+1)(x-1)}{x-1}

Cancel the factor (x - 1) on the numerator/ denominator, leaving

\frac{f(x)}{g(x)} = 2x + 1 with remainder 0 ≠ 2 ← False

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Therefore, they are on the 3rd number in specific group.

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\qquad \qquad \textit{inverse proportional variation} \\\\ \textit{\underline{y} varies inversely with \underline{x}} ~\hspace{6em} \stackrel{\textit{constant of variation}}{y=\cfrac{\stackrel{\downarrow }{k}}{x}~\hfill } \\\\ \textit{\underline{x} varies inversely with }\underline{z^5} ~\hspace{5.5em} \stackrel{\textit{constant of variation}}{x=\cfrac{\stackrel{\downarrow }{k}}{z^5}~\hfill } \\\\[-0.35em] ~\dotfill

\stackrel{\begin{array}{llll} \textit{\tiny "t"ime varies}\\ \textit{\tiny inversely with "s"peed} \end{array}}{t = \cfrac{k}{s}}\qquad \textit{we know that} \begin{cases} t=\stackrel{minutes}{40}\to \stackrel{hrs}{\frac{2}{3}}\\ s=\stackrel{m/h}{9} \end{cases} \implies \cfrac{2}{3}~~ = ~~\cfrac{k}{9} \\\\\\ 18=3k\implies \cfrac{18}{3}=k\implies 6=k~\hfill \boxed{t=\cfrac{6}{s}} \\\\\\ \textit{when s = }\stackrel{m/h}{12}\textit{ what is "t"?}\qquad t=\cfrac{6}{12}\implies t=\cfrac{1}{2}

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2 years ago
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