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Find the Distance between points (-2, -9) and (4, -3) to the nearest tenth.
Show work. (help please?)

We have the points P₁ and P₂ :

\(\begin{array}{|rcll|} \hline P_1 &=& (x_1, y_1) \\ &=& (4, -3) \\\\ P_2 &=& (x_2, y_2) \\ &=& (-2, -9) \\ \hline \end{array}\)

Formula to find the Distance between these Points is:

\(\begin{array}{|rcll|} \hline && \sqrt{(x_1-x_2)^2+ (y_1-y_2)^2} \\\\ &=& \sqrt{[4-(-2)]^2+ [-3-(-9)]^2} \\\\ &=& \sqrt{(4+2)^2+ (-3+9)^2} \\\\ &=& \sqrt{6^2+ 6^2} \\\\ &=& \sqrt{2\cdot 6^2} \\\\ &=& \sqrt{2}\cdot \sqrt{6^2} \\\\ &=& \sqrt{2}\cdot 6 \\\\ &=& 1.41421356237\cdot 6 \\\\ &=& 8.48528137424 \\ \hline \end{array}\)

The distance to the nearest tenth is 8.5

int (In x)^2/x  dx

\(\int\;\frac{(lnx)^2}{x}\;dx\\ \qquad Let\;\;u=lnx\\ \qquad \frac{du}{dx}=\frac{1}{x}\\ =\int\;u^2\frac{du}{dx}\;dx\\ =\int\;u^2\;du\\ =\frac{u^3}{3}+c\\ =\frac{(lnx)^3}{3}+c\\ \)

check

LHS

=ln ((1+sqrt(1+4e))/2) + ln(((1+sqrt(1+4e))/2)-1)

= 0.9999999999999998  

=RHS

In x + In(x-1)=1

x>0

x-1>0      x>1

\(ln[x(x-1)]=1\\ ln[x^2-x]=1\\ e^{ln[x^2-x]}=e^1\\ x^2-x=e\\ x^2-x-e=0\\ x=\frac{1\pm\sqrt{1+4e}}{2}\\ \text{but x>1 so}\\ x=\frac{1+\sqrt{1+4e}}{2}\\\)
 

Take the integral:
 integral e^x sqrt(e^x + 1) dx
For the integrand e^x sqrt(e^x + 1), substitute u = e^x + 1 and du = e^x dx:
 = integral sqrt(u) du
The integral of sqrt(u) is (2 u^(3/2))/3:
 = (2 u^(3/2))/3 + constant
Substitute back for u = e^x + 1:
Answer: |= 2/3 (e^x + 1)^(3/2) + constant

u= -2b+2a-3

Solve for a:
u = -3 + 2 a - 2 b

u = -3 + 2 a - 2 b is equivalent to -3 + 2 a - 2 b = u:
-3 + 2 a - 2 b = u

Add 2 b + 3 to both sides:
2 a = 3 + 2 b + u

Divide both sides by 2:
Answer: |a = 3/2 + b + u/2

Sorry, young lady!! I calculated your problem using 7.6% compounded monthly because the payments are monthly. However, it is very easy to convert 7.6% from copounded annually to compounded monthly. So, 7.6% annual compound = 7.35% monthly compound. So, based on this Rolfe's RRSP and his retirement payments will be as follows:

1) His RRSP balance after 30 years will be =$980,250.96

2) His monthly payments at retirement will be =$7,146.98

3) His last payment will be =$7,146.98 + 0.04 =$7,147.02, after which his balance will be zero.

rolfe set up an RRSP for a client whn the client was 30 years old . the client put $750 into the account every month until they retired at age 60 . the account averaged an annual rate of return of 7.6 % per year compounded annually . if the client then wanted to take a monthly disbursement from the account starting at age 60 for the next 25 years , what would the monthly disbursment be if the  RRSP investment retained the same terms and after 25 years woulf have a zero balance ?

1) This is the formula you would use to find the balance in his RRSP after 30 years, or when is 60.

FV=P{[1 + R]^N - 1/ R}=FV OF $1 PER PERIOD.

FV =750{[1 + 0.076/12]^(30*12) - 1 / (0.076/12)}

FV =$1,031,049.65 - this is what Rolfe will have in his RRSP in 30 years.

2) This is the formula you would use to calculate his monthly payments for 25 years.

PV=P{[1 + R]^N - 1*[1 + R]^-N}* R^-1=PV OF $1 PER PERIOD.

$1,031,049.65 = P*[1 + 0.76/12]^(25*12) -1 *[1 + 0.076/12]^-(25*12) * (0.076/12)], solve for P.

P =$7,686.56 - Rolfe's monthly payment for 300 months, or 25 years.

His last payment will be=$7,686.56 + $2.04 =$7,688.60 after which he will have zero balance.

it equals -500

what is u= -2b+2a-3 , Solve for A

An American reported that the diameter of the sun is approximately 8.656*10^5 miles. WRite this in kilometres, using scientific notation written correct to four significant figures. There are 1.609km in a mile. If the sun's diameter is 109 time that of the earth, what is the earth's diameter, correct to three significant figures?

1 mile =1.609344 km.

[8.656 x 10^5] x 1.609344 =1,393,048.166 km =1.393 x 10^6 km - diameter of the Sun.

1,393,048.166 / 109 =12,780.258 km =1.278 x 10^4 km - diameter of the Earth.

display the value of two different investments after 2 years , 6 years , 10 years , 14 years , and 17 years on the grid provided . investment one was $10000 earning simple interest at 5% per year . investment two was $10000 earning the same rate but compounded semi-annually .

Simple interest:

10,000 x 5% =$500 simple interest for 1 year.

$500 x 2 =$1,000 for 2 years.

$500 x 6 =$3,000 for 6 years.

$500 x 10 =$5,000 for 10 years.

$500 x 14 =$7,000 for 14 years.

$500 x 17 =$8,500 for 17 years.

Compound interest:

FV = PV[1 + R]^N

FV =10,000[1 +0.05/2]^(2*2) =$11,038.13 - $10,000 =$1,038.13 compound interest for 2 years.

$3,448.89 for 6 years.

$6,386.16 for 10 years.

$9,964.95 for 14 years.

$13,153.22 for 17 years.

If $5000 is deposited at the end of each quarter in an account that earns 6% compounded quarterly, after how many quarters will the account contain $60,000? (Round your answer UP to the nearest quarter.)

FV=P{[1 + R]^N - 1/ R}=FV OF $1 PER PERIOD.

60,000 = 5000{[1 + 0.06/4]^N -1}/ (0.06/4)

Solve for N over the real numbers:
60000 = 333333. (1.015^N - 1)

333333. (1.015^N - 1) = 1000000/3 ((203/200)^N - 1):
60000 = 1000000/3 ((203/200)^N - 1)

60000 = 1000000/3 ((203/200)^N - 1) is equivalent to 1000000/3 ((203/200)^N - 1) = 60000:
1000000/3 ((203/200)^N - 1) = 60000

Multiply both sides by 3/1000000:
(203/200)^N - 1 = 9/50

Add 1 to both sides:
(203/200)^N = 59/50

Take the logarithm base 203/200 of both sides:
Answer: |N = (log(59/50))/(log(203/200))=~11 quarters.

Here is a gigantic hint from AoPS Wiki (http://artofproblemsolving.com/wiki/index.php?title=Committee_forming)

FV={P{[1 + R]^N - 1/ R}} * [1 + R]

36,000 ={P x {[1.08]^10 - 1 / 0.08}} * 1.08

36,000 ={P x 14.486562 x 1.08}

36,000 ={P x 15.645487} divide both sides by 15.645487

P = 36,000/15.645487 = $2,300.98 amount that must be deposited at the BEGINNING of each year.

Here is how you can do it with Desmos but it is kind of a guess method that I dont think all that much of.

Still it would serve  the purpose ok I guess,   :)

Also, it might help you understand how the numbers affect the graph. 

4 = 4% x y

4 =0.04 x y divide both sides by 0.04

y =4 / 0.04 = 100 years for $1 to grow to $4 at simple interest of 4%.

P.S. compare this to 4% compounded annually at which you will quadruple your money in about 35 years.

What (regression) calculator are you using ?  Do you have one that you use for school ?

Here is an online explanation of how to do it on a   T1-83 and 84.

I have neither but I know many people do...

scott borrowed $3500 to cover the costs of taking a firefighter course in texas . his bank offered him a simple interest loan at a rate of 4.7% with single repayment in 6 months. another bank offered him a loan that was compounded monthly at an annual rate of 4.2% with a single repayment in 6 months . which option would cost him the least amount of interest and by how much less ?

Option @ 4.7% = 3,500 x 4.7% =$164.50 simple annual interest.

$164.50 / 2 =$82.25 this is what Scott will pay in 6 months at simple interest.

Option@ 4.2% =3,500 x (1 +0.042/12)^6 =3,500 x 1.0035^6=$3,574.15.

$3,574.15 - $3,500 =$74.15 this is what Scott will pay in 6 months at compounded interest.

The compound option is cheaper by: $82.25 - $74.15 =$8.10.

Thank you so much for all your help Melody!!

How do I do this?

How much must be deposited at the beginning of each year in an account that pays 8%, compounded annually, so that the account will contain $36,000 at the end of 10 years? (Round your answer to the nearest cent.)

The amount will grow by a factor 1.08 each year so you just have to solve this equation.

\(36000=X*1.08^{10}\)

The round up to the next cent.

Can you do that yourself ?

-13. Order of operations makes it 3 - (18) + 2 == -15 + 2 == -13

Ok NinjaDragon98 has got more questions...  It is always good to ask questions when you don't totally understand!    

There are 2 common ways to measure angles.  

One is in degrees where there are 360 degrees in a revolution

and the other in radians were there are 2pi radians in a revolution.

Yes it does make a difference what your graph trig functions in.

Say you are graphing sin(theta)

If it is in degrees then the wave length will be 360 (degrees)  

If it is in radians then the wave length will be  2pi  (radians)  = about 6.28

Of course   360 degrees = 2pi radans but the numbers on the graph will be in accordance with the units you are using.

All the graphs I have presented are in radians.When you are in Desmos, there is a little spanner on the right side.  If you click that you will see radians and degrees buttons at the bottome.  I have set my graphs to radians.

Whenever I talk about degrees in this question it is concerning temperature - NOT anglular measure.

Your logic for Q23 is fine.  The question suggests tht you draw the graphs and so if you answer from what you can see that is should be all that is required.

However the number in front of the sine function gives the amplitude of the wave.  The amplitude is the maximum distance from the middle.  Since 20.71 is the biggest amplitude Osaka will have the biggest temp variation.  It would be good for you to assimilate this into your understanding of sine and cosine graphs. :)

Q24. Yes your logic is fine. It is the same as mine. :)

I think that covers what you ahve asked so far...  Any more questions, yes? then keep asking.   

I do prefer you to ask your questions on the froum (on this thread if it is for this question) Then you can send me a note with the address stating that you would like more help :) 

If you do not message me I may not know that you have asked a question but I still would like that question to be asked on the forum :))