$\begin{cases}x^3\qquad\qquad\ \ \ when\ x\geq3\\5\qquad\qquad\quad \ when\ 0\ \textless \ x\ \textless \ 3\\x^2-x+2\quad\ when\ x\leq0 \end{cases}\\\\\\ (a)\\ f(0)\iff x=0\implies f(x)=x^2-x+2\\\\f(0)=0^2-0+2=2\\\\(b)\\ \{f(2)\iff x=2;\ \ f(1)\iff x=1\}\implies f(x)=5\\\\ f(2)=5\quad\wedge\quad f(1)=5\\\\ f(2)-f(1)=5-5=0$

$(c)\\ \big\forall \limits_{ n\in R}\ n^2\geq 0\ \ \Rightarrow\ \ \big\forall \limits_{ n\in R}\ (-n^2)\leq 0 \ \ \Rightarrow\ \ x\leq0\ \ \Rightarrow\ \ f(x)=x^2-x+2\\\\f(-n^2)=(-n^2)^2-(-n^2)+2=n^4+n^2+2$

5 0

4 years ago

The sine of the complimentary angle of 28° a. -0.86 c. -0.74 b. 0.88 d. 0.76

In-s [12.5K]

Step One
Find the complimentary angle of 28o Compliments add to 90o.

28 + c = 90
c = 90 - 28
c = 62o

Step two
Find the sin of 62o

sin(62) = 0.8829

So the answer is B

Another way to do it is to take the cos of 28
Cos(28) = 0.8829 which is still B.

6 0

3 years ago