All Answers 358134 Answers

Thanks Bertie,

I think i can see that.  I shall think on it :/

Thanks for the comments Melody.

Notice that the 6n plus or minus 1 thing is not unique, 2n or 4n plus or minus 1 share the same property.

I chose 6n because the 6 happened to fit in with the question.

$${\frac{{\frac{\left(-{\mathtt{5}}\right)}{{\mathtt{7}}}}{\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{3}}}{\left(-{\mathtt{8}}\right)}} = {\frac{{\mathtt{15}}}{{\mathtt{28}}}} = {\mathtt{0.535\: \!714\: \!285\: \!714\: \!285\: \!7}}$$

$${\frac{{\frac{{\mathtt{2}}}{\left(-{\mathtt{7}}\right)}}{\mathtt{\,\times\,}}{\mathtt{7}}}{{\mathtt{5}}}} = {\mathtt{\,-\,}}{\frac{{\mathtt{2}}}{{\mathtt{5}}}} = -{\mathtt{0.4}}$$

$${\frac{{\mathtt{2}}}{\left(-{\mathtt{7}}\right)}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{7}}}{{\mathtt{5}}}} = {\frac{{\mathtt{39}}}{{\mathtt{35}}}} = {\mathtt{1.114\: \!285\: \!714\: \!285\: \!714\: \!3}}$$  =   $$1\frac{4}{35}$$

The basic equation would be

y = [(x + 5)(x  - 1)] / [(x + 2) ( x + 6) ]

Note, however, that when x = 0, y = -5/12....but we need to have y = 2,  when x = 0.......this implies that, if we just multiipied the basic function by -24/5, this would produce what we want since (-5/12)(-24/5) = 2

So we have...

y = [-24(x + 5)(x  - 1)] / [5(x + 2) ( x + 6) ]  =  (-24x^2 - 96x + 120) / (5x^2 + 40x  + 60)

Write an equation for a rational function with: Vertical asymptotes at x = -2 and x = -6 x intercepts at x = 1 and x = -5 y intercept at 2

$$y=\dfrac{\left(-4.8\right)\left(x+5\right)\left(x-1\right)}{\left(x+6\right)\left(x+2\right)}$$

√150 = 5√6

√24 = 2√6

So √150 + √24 = 7√6

.

Average acceleration = -20/5 = -4 m/s²

Force = mass*acceleration  (Note that kg*m/s² will give a result in Newtons).

.

$$\\(-32)^{\frac{2}{5}}\rightarrow(-2)^2\rightarrow 4\\\\-(32)^{\frac{2}{5}}\rightarrow -(2)^2\rightarrow-4$$

.

You can use s = ut + (1/2)at² where s = distance, u = initial velocity, t = time and a = acceleration:

$${\mathtt{s}} = {\mathtt{0}}{\mathtt{\,\times\,}}{\mathtt{0.5}}{\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,\times\,}}{\mathtt{1.5}}{\mathtt{\,\times\,}}{{\mathtt{0.5}}}^{{\mathtt{2}}} \Rightarrow {\mathtt{s}} = {\mathtt{0.187\: \!5}}$$

s= 0.1875 metres   Not very far!

.

Write an equation for a rational function with: Vertical asymptotes at x = 4 and x = 6 x intercepts at x = -4 and x = 2 Horizontal asymptote at y = 9

y = [9(x+4) (x -2)] / [(x -4) (x -6)]   =  [9x^2 + 18x - 72]  / [x^2 - 10x + 24]

Using implicit differentiation, find the equation of the tangent line to the given curve at the given point: 3x2y2 − 3y −17 = 5x +14 at (1,−3)

I. implicit differentiation:

$$\small{\text{$
\begin{array}{rcl}
3x^2y^2 - 3y -17 &=& 5x +14 \\
f(x,y) &=& 3x^2y^2 - 3y-5x -31 = 0\\\\
f'(x,y) &=& -\dfrac{ \dfrac{ d( 3x^2y^2 - 3y-5x -31 ) } {dx} }
{ \dfrac{ d( 3x^2y^2 - 3y-5x -31 ) } {dy} }\\\\\\
f'(x,y) &=& -\dfrac{ 6xy^2-5 }
{ 3x^2\cdot 2y-3 }\\\\\\
f'(x,y) &=& \dfrac{ 5- 6xy^2 }
{ 6yx^2-3 }
\end{array}
$}}$$

II.  tangent line

$$\small{\text{$
\begin{array}{rcl}
x_p=1 \qquad y_p = -3\\\\
f'(x_p,y_p) &=& \dfrac{y-y_p}{x-x_p}\\\\
y &=& f'(x_p,y_p)\cdot(x-x_p)+y_p \qquad | \qquad f'(x_p,y_p) = \dfrac{5-6\cdot 1\cdot (-3)^2} {6\cdot (-3)\cdot 1^2 -3 } = 2.\overline{3}\\\\
\mathbf{y} &\mathbf{=}&\mathbf{ 2.\overline{3} \cdot(x-1)-3}\\\\
\mathbf{y} &\mathbf{=}&\mathbf{\dfrac{7}{3} \cdot(x-1)-3}
\end{array}
$}}$$

3x^2y^2 − 3y −17 = 5x +14

6xy^2 + 6x^2yy ' - 3y '  = 5

y' ( 6x^2y - 3)  =  5 - 6xy^2

y' =   ( 5 - 6xy^2) / (6x^2y - 3)        and the slope at (1, -3)  = [(5 - 6(1)(-3)^2] / [(6(1)^2(-3) - 3) ] = -49 /  -21   =   7/3

And the equation of the tangent line at this point is.....

y = (7/3)(x - 1) -3  

y =(7/3)x - (7/3) - 3

y = (7/3)x - 16/3

how do i simplify 2p^2+3p+1=0

$$\small{\text{$
\mathbf{2p^2+3p+1=2\left(p+\dfrac{1}{2}\right) \left( p+1\right)=0
}
$}}$$

how? haha 

U = 3/2

Oh dear MG,

We have stubbled on a concept that you absolutely need to understand.

Length measures distance. The units might be cm, m, yards, km etc

The perimeter or the length or bredth or height of am object is a length.

You can measure theses with a ruler or with a measure tape etc

Area measures how many SQUARES fit INSIDE a shape.  

So the units might be   $$cm^2\;\; or\;\;m^2\;\;or \;\;miles^2$$

The units are square units because they are counting squares.

Work through this page and think really hard about what you are being shown.

w = width
h = height
l = length
s = side

SA = wh + lw + lh + ls

SA = wh + lw + lh + ls

the answer is m=84