(b) $-+--3-=12$ $3$ $4$ $2$ $--+x-\dfrac {1} {2}$ $3$ 4. Solve : (a). $\dfrac {3x-2} {4}$ $\dfrac {x-1} {3}=\dfrac {5-x} {3}$ /16 (b)- $\dfrac {4x-3} {3}$ $\dfrac {3x+2} {5}$ $=\dfrac {7-2x} {3}+\dfrac {2x} {5}$ 115 1 5. Solve : (a) $\dfrac {5} {x+6}=\dfrac {1} {x-2}$ (b) $\dfrac {7} {x+9}=$ $\dfrac {-32} {15x-2}$ 6. Solve : (a) $\dfrac {x-3} {3x+2}=\dfrac {x+5} {3x+4}$ (b) $\dfrac {x-2} {2x-5}=\dfrac {x+4} {2x+1}$ Solve $\left(a\right)\left(x-4\right)^{2}+5=\left(x+2\right)^{2}-\left(9-14x\right)$ $\left(b\right)\left(x-2\right)^{2}+9=\left(2x+1\right)^{2}-\left(5x-6\right)$ Solve : (a) $\dfrac {2x-1} {5x+3}=\dfrac {2x+2} {5x+16}$ (b) $\dfrac {4x+1} {3x+2}=\dfrac {4x+6} {3x+7}$ Solve : (a) xx ++2 1 xx+-2 1 $=\dfrac {x+5} {x^{2}-1}$ (b) x 2-+ x 2 $\dfrac {2x-1} {x-2}=\dfrac {5x-1} {x^{2}-4}$

The is the statement of the Mean Value Theorem from your $te\times tb00k$ $Tneoren$ $m4.5$ $Ne8n$ Value Theorem Let f be continuous over $leclose$ interval $\left(a.b\right)aod$ $i$ $eo$ $xd$ $csnlc$ $\left(ab\right)$ $mmd$ exists at least one point cE $\left(a.b\right)$ $si1dhtba$ $f^{'}\left(c\right)=\dfrac {f\left(b\right)-f\left(a\right)} {b-a}$ What is the geometric interpretation of the conclusion of the theorem? O The tangent line to the graph of \(f\left(x\right)\) at $l\left(c1\right)$ is parallel to the secant line connecting \(\left(a,f\left(a\right)\right)\) and \(\left(b,f\left(b\right)\right)\). O The tangent line to the graph of $\left(f\left(lef\left(x\left(right\right)$ at \(c\) is the secant line connecting \(\left(a,f\left(a\right)\right)\) $ana$ \(\left(b,f\left(b\right)\right)\). O The tangent line to the graph $of$ $\left(f\left(leF\left(x\left(right\right)\left(\right)at1\left(c1\right)is$ perpendicular to the secant line connecting \(\left(a,f\left(a\right)\right)\) and A(\left(b,f\left(b\right)\right)\). O The tangent line to the graph of $\left(fleH\left(x\right)nght\right)\right)\right)$ $at$ $\left(c\right)\right)ishoizonta|$ $Tnere$ is more than one tangent line to the graph $0no$ $\left(flcf\left(x\right)night\right)\left($ $at1\left(c1\right)$

If the sum of two consecutive numbers is $45$ and one number is $X$ .This statement in the form of equation $1s:$ $\left(1$ Point) $\right)$ $○5x+1$ $1eft\left(x+1$ $r1gnt\right)=45s$ $○sx+1ef\left(x+2$ $r1gnt\right)=145s$ $sx+1x=45s$