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(f-g)x

= (5x-3)-(15x-6)

= 5x-15x-3+6

= -10x+3

A = x and A = 15?

7/8 = 0.875

1/4 = 0.25

7/8 - 1/4

= 0.875 - 0.25

= 0.625

Our guest 1 minute is not 6 degrees......

1 degree = 60 minutes

1 minute = 60 seconds....

Btw you can't type angles in with degrees minutes and seconds. You type it as degrees.

Example:

72°30'12''

= 72° + 30/60 ° + 12/3600 °

= \(\frac{21751°}{300}\)

\((a-b)^3 = a^3-3a^2b+3ab^2-b^3\)

Therefore\((r-5)^3=r^3-(3)(r^2)(5)+(3)(r)(5^2)-5^3=r^3-15r^2+75r-125\)

25 - 7 x 2 + 5 ÷ 5 = 12

Find the equation of the tangents to the circle x^2+y^2=9 with a slope=1

Implicitly differentiate

2x + 2yy'  = 0      divide through by 2

x + yy'  = 0

yy'  = -x

y'  = -x / y

If the slope is 1, then  y = -x       and we can use the equation to find the points where this occurs

x^2 + (-x)^2  = 9

2x^2  = 9

x^2  = 9/2

x = ± 3 / √2    so.....when  x = 3 / √2, y = -x =  - 3 / √2       and when x = -3 / √2, y = -x =  3 / √2

So.....we have two tangent line equations....the first is

y = (x - 3 / √2) - 3 / √2  →    y = x - 6 / √2 

And the second is

y = (x + 3/ √2 )   + 3 / √2    →   y = x + 6 / √2 

Here's a graph : https://www.desmos.com/calculator/vszg67cvt8

I have included the line y = x which also has a slope of 1 to show that our  two lines are parallel to this one... thus, all three lines have a slope of 1....!!!

This is the graph of a circle.......if you have had calculus, we can use implcit differentiation to find the slope of a line that is tangent to the circle at any point (x, y)......this is given by :

x^2+y^2+6x-10y+17=0           implicitly differentiate

2x + 2yy' + 6 -10y' + 0 = 0  

y' [ 2y -10 ]   =  - [ 2x + 6]      divide through by 2

y' [ y - 5]  = - [ x + 3]      divide by [ y - 5]

y'  = - [ x + 3] / [ y -5]       factor out a  -1

y' = [ x + 3] / [5 - y]         the right side is the slope of the tangent line for any (x , y)  on the circle.....so......you can use this slope and the point to write an equation for a tangent line just like you would in Algebra !!!........note that the slope is undefined when y = 5

p=10000 r=5 n=12 t=10 calculate compound intrest

FV=PV[1 + R]^N

FV=10,000[1 + 0.05/12]^(12*10)

FV=$16,470.09

P.S. Your question is not clear, so I have assumed a number of things: That 10,000=Present value, r=Interest rate, n=Compounding periods, and t=Number of years. If this is NOT accurate, then re-submit your question with clear explanation.

Simplify the following:
7/8-1/4

Put 7/8-1/4 over the common denominator 8. 7/8-1/4  =  7/8-(2)/8:
7/8-(2)/8

7/8-2/8  =  (7-2)/8:
(7-2)/8

7-2 = 5:
Answer: |  5/8

[1,000/4,000] x $840.55 =$210.14 cost per 1,000 sheets of board.

x^2 + y^2 = 9

Differentiate wrt x

2x + 2y*dy/dx = 0

Hence x + y = 0. (because dy/dx = 1), so y = -x

Put this in the original equation to get. 2x^2 = 9

Two solutions:  x = 3/sqrt(2) and x = -3/sqrt(2)

Corresponding values of y are y = -3/sqrt(2) and y = 3/sqrt(2) (because y = -x)

Equation of line with slope 1 is y = x + k where k is a constant.

Plug the known pairs of x and y into this equation to find the corresponding values of k.

loga(2x^2-7) = 2loga(x-1)

loga(2x^2-7) = loga(x-1)^2

2x^2 - 7 = (x - 1)^2

2x^2 - 7 = x^2 - 2x + 1

x^2 + 2x - 8 = 0

(x + 4)(x - 2)  = 0

Can you take it from here?

Does this help:

.

Like so:

\(\frac{dN}{dt}=-RNp\\ \frac{1}{N}\frac{dN}{dt}=-Rp\\\frac{d \ln N}{dt}=-Rp\\ \ln N - \ln N_0 = -Rpt\\ N=N_0e^{-Rpt}\)

Hello!

Thanks again...Ok so maybe this helps,

There was a loss realized on part of an activity in 2013 for 300,000€ and has not happened again.

The loss was related to certain activity that happened between 2010 and 2013.

A mitigating action was put in place so that such issue does not occur again. So risk for this is zero and probability zero.

However, a similar activity is running the risk of having the same issue.

It started in 2010 and has ran until 2015. In 2015 mitigating actions have been taken to stop the future issues, but the actions between 2010-2015 cannot be mitigated and the risk remains that a loss may occur for these.

The estimated risk is max 50mio-min 5 million (1mio per year)

We could say that both activities are the same so the issue has happened 1 in 6 years and 3 years ago.

Do you need more information?

Many thanks!

Solve what?

The distribution of actual weights of 8 oz. wedges of cheddar cheese produced at a dairy is Normal with mean 8.1 ounces and standard deviation 0.2 ounces. A sample of 10 of these cheese wedges is selected. What is the probability that the average weight of this sample is between 7.9 and 8.1 ounces?

the standard error of the mean is  

\(\bar x=8.1\\ SE_{\bar x}=\frac{\sigma}{\sqrt{n}}\\ SE_{\bar x}=\frac{0.2}{\sqrt{10}}\\ SE_{\bar x}=0.0632\\ \)

7.9 is  \(- \sqrt{10} = -3.162\) standard errors from the mean

and 8.1 is the mean

This is an area of more thant 3 standard errors  below the mean to the mean so the probability is greater than 99.7/2 = 49.85% and less than 0.5%

You could use tables if you want more accuracy.

If you have questions ask and I will try to answer you so that you understand.

How do I find an angle using tan, with the adjacent and the opposite side lengths?

If I have an adjacent length of 173 and a opposite length of 165?

\(tan\theta=\frac{opp}{adj}\\ tan\theta=\frac{165}{173}\\ \mbox{You are looking for an angle so you have to use inverse tan}\\ \theta=tan^{-1}\frac{165}{173}\\ \mbox{arc tan is the same as inverse tan so this goes in the web2 calc as}\\ atan(165/173)= \)

There is no question here 

This is a weird question - you are not even told what you have as the wording is "known and unknown"

In the first equation you do not have v but you have all the others so maybe you could use that

in the second equation you do not have initial displacement (usually considered to be 0 but maybe not here)

you might have everything else so maybe you could use that.    :///

Thaks Chris :)

 I believe this might be

3^(x+2)= 243         if so, we have that  243 = 3^5    ....so we can write

3^(x + 2)  = 3^5      we can just equate exponents, since the bases are the same

x + 2    = 5             subtract 2 from both sides

x  = 3

Let  m be the total distance he is from the skyscraper when he first sights it...and let h be the height of the skyscraper......so we have.....

tan (33) = h/m        →  m =    h/tan (33)

After he runs 350m....he is (m - 350)m from the skyscraper

And we have that

tan (52)  = h / [ m - 350]        substitute for m  to get an equation in one variable

tan (52)  = h / [ h/tan (33) - 350]          multiply both sides by  [ h/tan (33) - 350]  

[ h/tan (33) - 350]  * tan (52)   = h        simplify

h [ tan(52)/tan(33) ] - 350 * tan(52)   = h        rearrange  and factor

h [ tan (52) / tan(33)  - 1 ]  =  350 * tan(52)     divide both sides by [ tan (52) / tan(33)  - 1 ] 

h =  350 * tan(52)  / [ tan (52) / tan(33)  - 1 ]   ≈ about 461.39 m