Calculator search results

$4.$ $1tf\left(x\right)=3x^{2}-2^{2}-8$ and $h\left(x\right)=-4x^{3}+2x^{2}+6x$ find $f\left(x\right).b\left(x\right)$ $2$ $-x^{1}+4x^{2}-6x-8$ $c.$ $7x^{2}-6x-8$ $7x^{3}-4x^{2}-6x-8$ $b.$ $-x^{3}-4x^{2}-6x-8$ $4$

Using the \emph{removal of first derivative} method, the differential equation \( \frac{d^{2}y} $\left(d\times n$ $\left(2\right)\right)+P|ffac\left(dy\right)\left(dx\right)+Qy=F$ $dx\right)+Qy=RN\right)$ is transformed as \). For, the differential equation \frac{d^{2}y} $\left(d^{n}\left(2\right)y\right)$ $dx$ $\left(2\right)+2x$ $\left(0C\left(dy\right)\left(dx\right)+\left(x$ $2+1\right)y=\times n3+3x\right)$ the value of $\left(11\right)$

$∠$ $F$ From the sum $ot$ $x^{A}\left(2\right)-8$ with $3x-12$ subtract the sum $ofx-9$ with $3x-x^{A}\left(2\right)$ Multinlication takes place $\left(2\times A$ $\left(2\right)+5x+21\right)\left(x-$ $5\right)=$ #Please provide solution by using math formula.

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 $0no$ $\left(flcf\left(x\right)night\right)\left($ $at1\left(c1\right)$

$s|ef\left(-1n$ $\left($ }\right)^{50}\ $\right)$ \ | | is\ equal\ to\ $S$ $s1S$ $S-1S$ $s2S$ $s50s$

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) $\right)$ $○5x+1$ $1eft\left(x+1$ $r1gnt\right)=45s$ $○sx+1ef\left(x+2$ $r1gnt\right)=145s$ $sx+1x=45s$