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Anarel [89]
3 years ago
13

Help algebra 2 please

Mathematics
1 answer:
kati45 [8]3 years ago
7 0

Answer:

Step-by-step explanation:

(f*g)(x) = (-5x² + 2x + 7) (x +1)

          = x* (-5x² + 2x + 7) + 1*(-5x² + 2x + 7)

          = x*(-5x²) + x*2x + x*7 - 5x² + 2x + 7

        = -5x³ + 2x² + 7x - 5x² + 2x + 7

        = - 5x³ + <u>2x² -5x²</u>   <u>+ 7x + 2x </u>+7   {Combine like terms}

       =  -5x³ - 3x² + 9x + 7

4) (f*g)(x) = (x² + 2x + 4)(x - 2)

                = x*(x² + 2x + 4) - 2*(x² + 2x + 4)

                = x*x² + x*2x + x*4 - 2*x² - 2*2x -2* 4

                = x³ + 2x² + 4x  -2x² -4x - 8

                = x³ - 8

2) Speed=\dfrac{distance}{time}\\\\\\= \dfrac{d(h)}{t(h)}\\\\= \dfrac{2\sqrt{h}}{3\sqrt{4h}}=\dfrac{2*\sqrt{h}}{3*2\sqrt{h}}\\\\\\=\dfrac{1}{3}

1)d(h) = \sqrt{16h^{4}}=\sqrt{2*2*2*2*h*h*h*h}=2*2*h*h=4h^{2}\\\\\\t(h) = 3h^{4}-3h^{3}-5h^{2}= h^{2}(3h^{2}-3h - 5))\\\\Speed=\dfrac{4h^{2}}{h^{2}(3h^{2}-3h-5)}\\\\=\dfrac{4}{3h^{2}-3h-5}

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Solve the trigonometric equation:

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Make a substitution:

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Substitute back for  t = cos x:

     \begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}


Therefore,

     \begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}

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Solution set:   

\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}


I hope this helps. =)

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