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2 years ago

7. Find the exact value of a) cos (195) b) csc (75) Make sure to show each step

Mathematics

1 answer:

Ket [755]2 years ago

6 0

7.7777777777777777777777777777777777

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2 years ago

4x2 + 4x - 1, evaluate and fully simplify each of the - For the function f(0) following f(s + h) f(x + h) - f(5) h 10

Andrew [12]

Given the function f(x);

$f(x)=-4x^2+4x-1$

Evaluating the function f(x+h);

$\begin{gathered} f(x+h)=-4(x+h)^2+4(x+h)-1 \\ f(x+h)=-4(x^2+2xh+h^2)^{}+4(x+h)-1 \\ f(x+h)=-4x^2-4h^2-8xh^{}+4x+4h-1 \end{gathered}$

So;

$f(x+h)=-4x^2-4h^2-8xh^{}+4x+4h-1$

Evaluating the second function;

$\begin{gathered} \frac{f(x+h)-f(x)}{h}=\frac{-4x^2-4h^2-8xh^{}+4x+4h-1-(-4x^2+4x-1)}{h} \\ \frac{f(x+h)-f(x)}{h}=\frac{-4x^2-4h^2-8xh^{}+4x+4h-1+4x^2-4x+1}{h} \\ \frac{f(x+h)-f(x)}{h}=\frac{-4x^2+4x^2-4h^2-8xh^{}+4x-4x+4h-1+1}{h} \\ \frac{f(x+h)-f(x)}{h}=\frac{-4h^2-8xh^{}+4h}{h} \end{gathered}$

simplifying further;

$\begin{gathered} \frac{f(x+h)-f(x)}{h}=\frac{-4h^2-8xh^{}+4h}{h}=-4h-8x+4 \\ \frac{f(x+h)-f(x)}{h}=-4h-8x+4 \end{gathered}$

Therefore, we have;

$undefined$

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6 months ago