All Answers 358516 Answers

Wed 28/1/15

Thanks Chris :)

Very nice, Melody.....this one is a definite "Daily Wrap" candidate....!!!!

see if this page helps 

v=1/0.00006

now you can do it on the calc on the home page :)

First of all.....never trust anyone who can't spell "calculator" correctly.

Secondly......no one is forcing you to stay here....

Thirdly.....in the American South, we have a saying about your exit....

"Don't let the door hit ya' where the dog shoulda' bit ya'."  ....

T = D / R  → 50m / (8m/s) = 6.25 sec

A cylindrical can has a radius of 3.5cm and a volume of 416cm cubed. work out the height of the can give your answer correct 2  decimal place

$$V = \pi r^2 h \\\\
h = \dfrac{V}{\pi r^2} \quad | \quad r = 3.5\ cm \quad V=416\ cm^3 \\\\
h = \dfrac{416}{\pi (3.5)^2} = 10.81 \ cm \\$$

No prob....

$1200[{[1 + (.07/12) ] 180 – 1} / (.07/12)] = ?

Do you mean:

$$\small{\text{
$r = \frac{7}{12}\ \% $
and after
$n = 180 $
months, the sum
$
s = \$ \ 1200 * \left( \dfrac{ (1+r)^n - 1 } { r } \right)
$
}}\\
\small{\text{
$
s = \$ \ 1200 * \left( \dfrac{ \left( 1+ \dfrac{0.07}{12} \right)^{180} - 1 } { \dfrac{0.07}{12} } \right)
= \$ \ 1200 * \left( \dfrac{ 2.84894673087 - 1 }{ 0.00583333333 }\right)
$
}}\\\\
\small{\text{
$
= \$ \ 1200 * 316.962296721 = \$ \ 380354.76
$
}}$$

thank you for that answer

The equation of C is   $$x^2+y^2-8x-20y+100=0$$    with cetre(4,10)

given that  the slope and the x-intercept of a straight B are 1 and K respectively .If L cuts C at P and Q , find the ccoordinates of the mid -point of PQ in terms of K 

 --------------------------------------------------------------------------

 I am going to use the fact that the radius of a circle that bisects a chord is prependicular to the chord.

now the equation of the line is    y=1x+K

 -------------------------------------------------------

 So I want the equation of the line prependicular to  y=1x+k   through  (4,10)

the gradient will be -1

$$\\-1=\frac{y-10}{x-4}\\\\
-1(x-4)=y-10\\
-x+4=y-10\\
y=-x+4+10\\
y=-x+14$$

So now I want to find the intersection of  y=-x+14 and  y=x+k

$$\\-x+14=x+K\\
-2x=K-14\\
x=\frac{14-K}{2}\\\\
y=\frac{14-K}{2}+K\\\\
y=\frac{14-K}{2}+\frac{2K}{2}\\\\
y=\frac{14+K}{2}\\\\
\left(\frac{14-K}{2},\;\frac{14+K}{2}\right)$$

check    

$$\\-x+14=\frac{K-14}{2}+\frac{28}{2}=\frac{K+14}{2}\\$$      Great it seems to work.

I went all around the world with this question before I got to the point.

It really does pay to think about problem questions before you barge head long into them.  LOL

The area of the cross-section is just  667.5 mm^3 / 7.5 mm = 89 mm^2

(-5,-5) parallel to 3x+5y=8

Parallel line through Point (-5,-5) :  3x+5y= 3*(-5) + 5*(-5)    3x+5y= -15-25    $$\boxed{ 3x+5y= -40 }$$

(3C3 + 4C3 + 8C3)/15C3

$${\frac{\left({\left({\frac{{\mathtt{3}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{4}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{4}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}{\mathtt{\,\small\textbf+\,}}{\left({\frac{{\mathtt{8}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{8}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}\right)}{{\left({\frac{{\mathtt{15}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{15}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)}}} = {\frac{{\mathtt{61}}}{{\mathtt{455}}}} = {\mathtt{0.134\: \!065\: \!934\: \!065\: \!934\: \!1}}$$

There you go, Alan's answer and my answer look different but they work out the same because the 3! in mine cancels out.  

How about that!      

Assuming that all 8 workers start and stop at the same time,  8 shirts are produced in 12.5 minutes......then 1 shirt is produced every (12.5 / 8) = 1.5625 min

Then 1200 of them would take 1200 x 1.5625 minutes = 1875 minutes = 31.25 hrs...

limit x to infinity 2x+3/5x+7

$$\lim\limits_{x \to \infty} \dfrac{ 2x+3 }{ 5x+7 } \\\\
=\lim\limits_{x \to \infty} \dfrac{ x \left( 2+ \dfrac{3}{x} \right) }{ x \left( 5+ \dfrac{7}{x} \right) } \\ \\
=\lim\limits_{x \to \infty} \dfrac{ \left( 2+ \dfrac{3}{x} \right) }{ \left( 5+ \dfrac{7}{x} \right) } \quad | \quad \lim\limits_{x \to \infty} \left(\dfrac{3}{x} \right) = 0 \qquad \lim\limits_{x \to \infty} \left(\dfrac{7}{x} \right) = 0\\ \\\\
= \dfrac{2+0}{5+0} = \dfrac{2}{5}$$

If this is (2x+3)/(5x+7)...the limit of this as x approaches infinity is just (2/5)....remember..... we have a "same over the same" condition, here.....which means that, in a rational function, when the polynomials in the numerator and the denominator are of the same degree.....the function has a limit as x → ∞  = the ratio of the coefficients on the highest power of each polynomial.....

Here's a graph......

(4x^2+3)=60

$$\\(4x^2+3)=60\\\\
4x^2+3=60\\\\
4x^2=60-3\\\\
4x^2=57\\\\
x^2=\frac{57}{4}\\\\
x=\pm\sqrt{\frac{57}{4}}\\\\
x=\pm{\frac{\sqrt{57}}{2}}\\\\$$

there are 3*24=72 lots of 20 minutes in a day    It is one day isn't it?

The beginning and the end count to n=73

1,2,4,6,8,.....

$$\\T_n=ar^{n-1}\\\\
T_{73}=1*2^{72}\\\\$$

$${{\mathtt{2}}}^{{\mathtt{72}}} = {\mathtt{4\,722\,366\,482\,869\,645\,213\,696}}$$

13/4 miles

inverse trig is the same as arc trig

so use acos, asin and atan       You can press [2nd] to bring them up     

s=a/(1-r)

2009=328/(1-r)

and you can solve it yourself :)

well it would take 8 workers  12.5/8 minutes to make one shirt - If it really worked that way.

so

it would take them 12.5/8*1200 minutes to make 1200 shirts

2/3 (9-m)= 2m+7 

multiply both sides by 3 first

2(9-m)=3(2m+7)

now expand the brackets 

and can you take it from there?