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2 years ago

I need help with the questions

Mathematics

1 answer:

riadik2000 [5.3K]2 years ago

8 0

Answer:

Step-by-step explanation:

Hello

please find attached the graph

you can notice that $x^2-6x+9 = (x-3)^2$

(a) from the graph we can say that there is no maximum and one minimum in x = 3

and then f(3) = 0 so o is the minimum of f

(b) vertex is (3,0)

(d) as $x^2-6x+9 = (x-3)^2$ the root is 3

(e) y=9 for x = 0 so the y-intercept is 9

do not hesitate if you need further explanation

if you like my answer, tag it as the brainliest :-)

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$\bf \textit{Sum and Difference Identities} \\\\ sin(\alpha - \beta)=sin(\alpha)cos(\beta)- cos(\alpha)sin(\beta)$

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$\bf cos(A)=\cfrac{\stackrel{adjacent}{3}}{\underset{hypotenuse}{5}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{5^2-3^2}}\implies b = \pm 4 \\\\\\ \stackrel{IV~Quadrant}{b = -4}\qquad \qquad sin(A)=\cfrac{\stackrel{opposite}{-4}}{\underset{hypotenuse}{5}} \\\\[-0.35em] ~\dotfill\\\\ cos(B)=\cfrac{\stackrel{adjacent}{12}}{\underset{hypotenuse}{13}}\qquad \qquad \stackrel{\textit{getting the opposite side}}{b=\pm\sqrt{13^2-12^2}}\implies b = \pm 5$

$\bf \stackrel{IV~Quadrant}{b = -5}\qquad \qquad sin(B)=\cfrac{\stackrel{opposite}{-5}}{\underset{hypotenuse}{13}} \\\\[-0.35em] ~\dotfill\\\\ sin(A-B)=\cfrac{-4}{5}\cdot \cfrac{12}{13}-\left( \cfrac{3}{5}\cdot \cfrac{-5}{13} \right)\implies sin(A-B)=\cfrac{-48}{65} - \left( \cfrac{-15}{65} \right) \\\\\\ sin(A-B)=\cfrac{-48}{65} + \cfrac{15}{65}\implies sin(A-B)=\cfrac{-33}{65}$

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