13. solve for x a. 5 5^x+1-5 14. Show that the period of f(x)=cos(x) is 2pi 15. Solve the following in equality a. 3x-6lt 2(3-x)+5 C. 2(5-x)geqslant -2x+5 b (x^2-5x+6)(x^2-4)geqslant 0 d. 3x^2+8x+5gt 0 16. Suppose (gcirc f)(x)=4x^2-9 and g(x)=x^2-6x find f(x) 17. Which of the following is a relations are function or not?how? a. R= (x,y):vert xvert =vert yvert b. R= (x,y):x=vert yvert C. R= (x,y)=vert xvert -vert yvert =9 18. Let f(x)=3x+4 and g(x)=6x^2+15x-11 then determine a. fcirc g (g(2) f +g 1.9 (g)/(f) gof 19. Let f(x)=4x+5 and g(x)=x^2+2 then find f+g, f.gf/g

13. To solve for x in the equation $5^{x+1}-5=0$, we can rewrite it as $5^{x+1}=5$. Taking the logarithm of both sides, we get $(x+1)\log_5(5)=\log_5(5)$, which simplifies to $x+1=1$. Therefore, $x=0$.<br /><br />14. To show that the period of $f(x)=cos(x)$ is $2\pi$, we can use the fact that the cosine function repeats its values every $2\pi$ radians. This means that for any value of x, $cos(x+2\pi)=cos(x)$. Therefore, the period of $f(x)=cos(x)$ is $2\pi$.<br /><br />15. <br />a. To solve the inequality $3x-6<2(3-x)+5$, we can simplify it to $3x-6<6-2x+5$, which further simplifies to $5x<15$. Dividing both sides by 5, we get $x<3$.<br /><br />b. To solve the inequality $(x^2-5x+6)(x^2-4)\geq0$, we can factor the expression to $(x-2)(x-3)(x+2)(x-2)\geq0$. The critical points are $x=-2$, $x=2$, $x=3$, and $x=6$. Testing the intervals between these critical points, we find that the inequality is satisfied for $x\leq-2$ or $x\geq3$.<br /><br />c. To solve the inequality $2(5-x)\geq-2x+5$, we can simplify it to $10-2x\geq-2x+5$, which further simplifies to $10\geq5$. This inequality is always true, so the solution is all real numbers.<br /><br />d. To solve the inequality $3x^2+8x+5>0$, we can use the quadratic formula to find the roots of the corresponding quadratic equation $3x^2+8x+5=0$. The roots are $x=-1$ and $x=-\frac{5}{3}$. Testing the intervals between these roots, we find that the inequality is satisfied for $x<-\frac{5}{3}$ or $x>-1$.<br /><br />16. Given $(g\circ f)(x)=4x^2-9$ and $g(x)=x^2-6x$, we can find $f(x)$ by substituting $g(x)$ into $(g\circ f)(x)$. This gives us $f(x)^2-6f(x)=4x^2-9$. Solving this quadratic equation for $f(x)$, we get $f(x)=2x+3$ or $f(x)=-2x-3$.<br /><br />17. <br />a. $R=\{(x,y):|x|=|y|\}$ is a relation, but it is not a function because for each x, there are two possible values of y (one positive and one negative).<br /><br />b. $R=\{(x,y):x=|y|\}$ is a relation, and it is a function because for each x, there is only one possible value of y.<br /><br />c. $R=\{(x,y):|x|-|y|=9\}$ is a relation, but it is not a function because for each x, there are two possible values of y (one positive and one negative).<br /><br />18. <br />a. $f\circ g(g(2))=f(g(2))=f(6(2)^2+15(2)-11)=f(31)=3(31)+4=97$<br /><br />b. $f+g(1.9)=f(1.9)+g(1.9)=3(1.9)+4+6(1.9)^2+15(1.9)-11=7.7+4+13.86+28.5-11=62.16$<br /><br />c. $\frac{g}{f}(gof)=\frac{g(gof)}{f(gof)}=\frac{g(f(gof))}{f(gof)}=\frac{g(f(6(2)^2+15(2)-11))}{f(6(2)^2+15(2)-11)}=\frac{g(f(31))}{f(31)}=\frac{g(97)}{f(97)}=\frac{6(97)^2+15(97)-11}{3(97)+4}=\frac{56406+1455-11}{295}=\frac{57050}{295}=193.22$<br /><br />19. <br />a. $f+g(x)=f(x)+g(x)=4x+5+x^