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Think about it this way:

The front of the train must travel 500m for the train to completely clear the bridge....so

Distance/Time  = Rate   ...    and we have

500m/30sec  = 50/3m/s  = 3600(50/3)m/hr = 60000m/hr   =  60km/hr

write the equation of the line tangent to the circle x^2 +y^2=169 at the point (-5,12)

This is a circle of radius 13 that is centered at the origin

The slope of the line segment drawn from the circle's center to the point (-5/12) = the radius   = -12/5

And since a tangent line to a circle at a given point is always perpendicular to the radius at that point, the slope of the tangent line =    5/12

And the equation of the tangent line is as follows:

y - 12= (5/12)(x -  -5)    simplify

y - 12   = (5/12)x + 25/12

y = (5/12)x + 169/12

Here's a graph :  https://www.desmos.com/calculator/xefjptaopf

Find the equation of the line perpendicular to and containing the midpoint of the segment joining (7,-20) and (3,10)

The midpoint  =  [ (7 + 3)/2, (-20 + 10)/2 ]  =  [ 10/2, -10/2] =  [5, -5]

The slope of the line joining the two points is given by :

[10 - -20] / [ 3 - 7 ] =  [30/-4]  = [-15/2]

So....a line perpendicular to the one joining the two given points wil have a slope = 2/15.......and the equation is :

y - -5  =  (2/15)(x - 5)   simplify

y + 5   = (2/15)x - 2/3

y = (2/15)x - 17/3

Here's the graph : https://www.desmos.com/calculator/nuwyd9uvqe

the first term of a geometric series is 2, the nth term is 486 and the sum of the n terms is 728. find r and n.

\(a=2 \qquad T_n=486\qquad S_n=728\\~\\ ar^{n-1}=486 \qquad \frac{a(r^n-1)}{r-1}=728\\ 2*r^{n-1}=486 \qquad \frac{2(r^n-1)}{r-1}=728\\ r^{n-1}=243 \qquad \frac{(r^n-1)}{r-1}=364\\ \)

\(\frac{r^n}{r}=243 \qquad \qquad r^n-1=364r-364\\ r^n=243r \qquad\qquad r^n=364r-363\\ 243r=364r-363\\ 363=121r\\ r=3\\~\\ 3^n=243*3\\ 3^n=3^6\\ n=6\\ so\\~\\ n=6\qquad and \qquad r=3\)

the first term of a geometric series is 2, the nth term is 486 and the sum of the n terms is 728. find r and n.

First term=2

The ratio =3

The 6th and the last term=486

Sum of G.P =728

what is the product of (3-4i) and (7-3i)?

\((3-4i)(7-3i)\\ =3(7-3i) -4i(7-3i)\\ =21-9i -28i+12i^2\\ =21-37i+\;\;\;\;12*-1\\ =21-37i\;\;\;\;-12\\ =9-37i\\\)

if i make £141.10 a week how long would it take for me to make £77,000

77000/141.10 = 545.7122608079376329  weeks

545.7122608079376329/52 = 10.4944665539988006327 years

that is 10 and a half years :)

Since there are no brackets, this reads as 

9-3/1 over 3 + 1 =

9-(3/1) over 3 + 1

= 9-  3/3 +1

= 9 -  1  +1

=9

Which best compares the slopes and y-intercepts of the linear functions f and g, where f = 1/3x + 3 and g is shown in the table below?

x1234

g(x)36912

Is this part of a multiple choice question?

the gradient of f(x) is 1/3      the gradient of g(x) is 3         That is the positive reciprocal       f(x)=(x/3)+3

The y interecept of f(x) is 3    and the y intercept of g(x) is 0                                                  g(x)=3x

I sam not really sure what answer is being sort here.   

Here is what the graphs look like

An equation

\(|\frac{1}{x}-\frac{2}{x+1}|=\frac{2}{x} \qquad \qquad x\ne 0\)

How many real roots does it have:

a)Exactly 3 real roots

b)Exactly 2 real roots

c)Exactly 1 real root

d)It doesn't have any real roots

\(\begin{array}{rcll} |\frac{1}{x}-\frac{2}{x+1}| &=& \frac{2}{x} \qquad & | \qquad \text{square both sides} \\ \left(\frac{1}{x}-\frac{2}{x+1} \right)^2 &=& \left(\frac{2}{x} \right)^2 \\ \left(\frac{x+1-2x}{x(x+1)} \right)^2 &=& \frac{4}{x^2} \\ \left(\frac{ 1- x}{x(x+1)} \right)^2 &=& \frac{4}{x^2} \\ \frac{ (1- x)^2}{x^2(x+1)^2} &=& \frac{4}{x^2} \qquad & | \qquad \cdot x^2 \qquad \text{solution }~ x = 0 ~ \text{impossible}\\ \frac{ (1- x)^2}{ (x+1)^2} &=& 4 \\ 4\cdot(x+1)^2 &=& (1- x)^2 \\ 4\cdot( x^2+2x+1) &=& (1-2x+x^2) \\ 4x^2 + 8x+4 &=& 1-2x+x^2 \\ 3x^2 + 10x+3 &=& 0\\ \end{array} \)

\(\begin{array}{|rcll|} \hline 3x^2 + 10x+3 &=& 0 \qquad \qquad a = 3 \qquad b=10 \qquad c = 3 \\\\ x_{1,2} &=& \frac{ -10 \pm \sqrt{10^2-4\cdot 3 \cdot 3} } { 2\cdot 3 } \\\\ x_{1,2} &=& \frac{ -10 \pm \sqrt{100-36} } { 6 } \\\\ x_{1,2} &=& \frac{ -10 \pm \sqrt{64} } {6 } \\\\ x_{1,2} &=& \frac{ -10 \pm 8 } {6 } \\\\ x_{1} &=& \frac{ -10 + 8 } { 6 } \\ x_{1} &=& \frac{ -2 } { 6 } \\ \mathbf{ x_{1} }&\mathbf{=}& \mathbf{ - \frac13 } \qquad &| \qquad \text{solution impossible }~ x > 0 \\\\ x_{2} &=& \frac{ -10 - 8 } { 6 } \\ x_{2} &=& \frac{ -18 } { 6 } \\ \mathbf{ x_{2} }&\mathbf{=}& \mathbf{ - 3 } \qquad &| \qquad \text{solution impossible }~ x > 0 \\\\ \hline \end{array}\)

d) It doesn't have any real roots

I don't think so...

can't you round by yourself? What's the number you need rounded?

Oh, thats right my formula was for annual interests

FV=PV [ 1 + R]^N, Where R=Interest rate per period, N=number of periods, PV=Present value, FV=Future value.

FV=1,000[1+.02]^365

FV=1,000[1.02]^365

FV=1,000 x 1,377.40829

FV=$1,377,408.29

$1,377,408.29 - $1,000 =$1,376,408.29 Interest earned for 1 year @ 2% compounded daily.

In this case the formula is simply FV(n) = $1000*1.02^n,  where n is the number of days

(2% per day is very large! Guest#1's formula would apply if the interest were 2% per annum)

I don't think this is the formula that I am looking for. Because by simply doing the math for the first 5 days, it goes like this:

1000 + 1000*0.02 = 1020, 1020 + 1020*0.02 = 1040.4, 1040.4 + 1040.4*0.02 = 1061.2 .....

With the formula that you proposed to me in 365 days the total profit is only 20.2, while in my calculations above in 3 days(!) I already have 61.2 profit. So which is the formula exactly?

Thanks

Hi, to calculate that you can use this formula :  

Final Money = A₀  * ( 1 + r/n )^nt 

nt = the time period in days if it is 3 years it would be 3*365;

r = interest; 6 % percent would be : 0.06

n = 365 ; if you want to calculate the profit daily

From my course on Calculus :)

Hope it helps,

Use the calculator on the home page here to do the calculations.

f(3x) = (3x)^2 - 3*(3x) + 8 → 9x^2 - 9x + 8

i need the answer

Volume = pi*r^2*h/3

r = radius, h = height

.

E = 0.6*12000 + 0.2*0 - 0.2*6000 → 6000

Not quite Asinus!  Slope = (y2- y1)/(x2 - x1)

So, slope = (3-7)/(9-4) = -4/5

Hence, slope of perpendicular = 5/4

If the recipie Liam is using requires 3 cups of flour, and he only has 2, then he has 2/3 of the amount in the recipie. This means he would need to use the same proportions of sugar as flour - 2/3. Therefore he needs to use 2/3 x 1 of the amount in the recipie, which is two thirds (2/3) of a cup.

Hello Guest!

What is the slope of a line that is perpendicular to

the line that passes through the points P1 (4, 7) and P2 (9, 3)?

m1 \( = \frac{y2 -x2}{y1-x1} \)

\(m1 = \frac{3- 9}{7- 4} \)

m1 = - 2

The slope of the perpendicular function is the

negative reciprocal of the slope of the output function.

m2 = - 1 / m1

\(m2 = \frac{- 1}{-2} = \frac{1}{2} \)

The slope of the perpendicular function is

\(m2 = \frac{1}{2} \)

Greeting asinus :- )

   !

I am not sure that the syntax is accurate. If the first equation should be

\(f(x)=({1\over2})^{2x}\)

then it is not linear. 

For x=0, f=1

For x=1, f=1/4

For x=2, f=1/16

If the second equation should be

\(f(x)=-{1\over{2}}*(x-2)\)

Then you can simplify to \(f(x)=-{x\over{2}}+1\)

which is linear. See that

For x=0, f=1

For x=1, f=0.5

For x=2, f=0

EDIT: I forgot the minus there, whoops.