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Oh. Sorry about that, 3 people.

THANK YOU SO MUCH! I really appreciate it! :)

Mr. Kheung had 2,000 dollars to invest. He invested part of it at 5% per year and the remainder at 4% per year. After one year, his investment grew to 2,095 dollars. How much of the original investment was at the 5% rate?     

Assuming simple interest

Let the amount invested at 5% per year be represented by X 

Let the amount invested at 4% per year be represented by 2,000 – X  

The total interest is the final value less the principal which is 2,095 – 2,000 = 95

The total interest is the sum of the two investments  

(0.05)(X) + (0.04)(2,000 – X)  =  95  

Can you take it from there?  Solve for X.  Go ahead and try. and use the following to check your answer.  

                                           (0.05)(X) + (0.04)(2,000 – X)  =  95  

                                                              .05X + 80 – .04X  =  95     

Combine the X's to get .01X  

and subtract 80 from each side                                 .01X  =  15  

Divide both sides by .01                                                  X  =  1500  invested at 5%   

_.

40890/47*34*15*20/6/50*11*8*7/17/33+46595=

Доброго времени суток уважаемый гость!

Вычисление точек (умножение, деление) перед вычислением линий (сложение, вычитание).

Вычислите точки в порядке слева направо.

\(40890/47*34*15*20/6/50*11*8*7/17/33\)       \(+\ 46595\)

 |                         Расчет баллов                                  |Расчет линии|            |              

                           32480                                                           +  46595    =   79075    

в дружбе

  !            

The slope of a line is -2 and its x-intercept is (5,0). What is the y-intercept point of the line? Express your answer as an ordered pair.

Point slope form of the line would be

(y-0) = -2 (x-5)   

re-arranged into y = mx+b form      m = slope   b = y axis intercept   

y = -2x+10 

I magnified the image size on my monitor, and, although the larger the size, the less the resolution, I'm pretty sure that the green angle is 39^o so, Guest OP, if you know that the angle actually is 39^o simply substitute 39 for the 30 in Melody's subtraction. 

_.

*** edited question ***

John buys 5 shirts and 3 trousers from a garment store at ₹900.  In the following week, when the store offers a discount of 10%

 on the same brand of the shirt and a discount of 25% on the same brand of the trouser, Jacob buys 5 shirts and 2 trousers for ₹630. Find the price at which John buys a trouser.

5 s + 3 t = 900                 10% discount = 90% price    25% discount = 75% of original price

5 s .9   +   2 t*.75 =630  

 4.5s    +  1.5 t = 630     two equations (in red) , two unknowns ....do you know how to solve?

First, ignore the fact that there is a frog, and just calculate the number of ways that Marvin can reach \((5, 7)\) only going up and right. 

To do that, Marvin must move 5 units to the right and 7 upwards to the right using exactly 12 moves. This is the same question as all the different distinguishable combinations of the string of letters \(RRRRRUUUUUUU\), where R represents right and U represents up. There are \({12 \choose 5} = {12 \choose 7} = 792 \) ways to do that.

Now, we need to find the number of ways that Marvin will encounter the frog. Specifically, Marvin needs to move 4 units to the right and 3 units upwards in his first 7 moves (while his other 5 moves could be in any order). Using the same logic above, the number of possible different distinguishable ways to order the string of letters \(RRRRRUUUUUUU\) in which the first 7 letters must contain 4 R's and 3 U's, is equal to \({7 \choose 3} \cdot {5 \choose 1} = 175\) ways. That is the number of ways that Marvin will meet the frog and get eaten.

Since we don't want that to happen, just subtract the total number of ways he is going to get to (5, 7) by the number of ways he is going to get eaten, which is equal to \(792-175=\boxed{617}\) ways. 

OK, got it   (duh!!!!)

Thanks, Melody   !!!!

I get it Chris, it is because BB'  and BC are both tangents and BO goes through the centre of the circle, so

angle  CBO and angle OBB' have to be congruent.

You could use the 3 equal sides test of congruency. 

Plus if two tangents intersect then the distances from the cross point to the circle intersections will be congruent.

.....

I still have a question,

How do you know that   

angle CBO = angle OBB'      ?....

Yeah.....I  was wondering about that too....

Thanks Tiggsy,

I still have a question,

How do you know that   

angle CBO = angle OBB'      ?

Here is the pic to go with Tiggsy's explanation.

It is interactive, you can grab the point T and move it around to see that the result says the same.

The height of the trapezoid =  the diameter of the  circle =   4.8 cm

Call the  sum of the  bases  , S

So  

24  = (1/2) (4.8)  (S)

24  =  2.4  S

24 /2.4 =  S  =   10

And BC  = 1/2 S  =  5   =  AD

So.....the perimeter  =   S  + 5  +  5 =    10 + 5   + 5    = 20 cm

I suspect Tiggsy meant the following for B' (with an analogous point for C'):

Thanks Chris,

It is the rates questions, that you seem to do effortlesslyy, that always confuse me  

Don't attempt to use a calculator here!

180-105 = 75

I am not sure if that green angle is 39 or 30. I will assume 30

180-30-75=75

x is vertically opposite the angle that I found so it is 75 degrees too.

Nice work, Melody......these nested root  things  always  frustrate me........you made that look  TOO  easy....LOL!!!

\sqrt{2021\sqrt{2021\sqrt{2021\sqrt\cdots}}}

Let

  \(y=\sqrt{2021\sqrt{2021\sqrt{2021\sqrt\cdots}}} \\ then\\ y=\sqrt{2021y}\\ y^2=2021y\\ y^2-2021y=0\\ y(y-2021)=0\\ y=0 \;\;or \;\;2021\\ y\ge0\\ so\\ y=2021 \)

Good work, Tiggsy......could   you provide a sketch   of  what you did   ???

I think there is a  more ingenious  way to solve this.....but....by brute force  ......

x + 1/x   = sqrt (3)

x^2  + 1  =  sqrt (3)  x

x^2  - sqrt (3) x   =  -1

x^2  -  sqrt (3) x  +  3/4  =  -1 + 3/4

(x  + sqrt (3) /2) ^2   =  -1/4        {take the  positive  root  since  x + 1/x   is positive}

x - sqrt (3)/2  =      (1/2) i

x =   sqrt (3/2)  +  (1/2)i

x^3   =   (sqrt (3)/2  + (1/2) i)^3  = 

((1/2)  (sqrt (3)  + i)^3       {binomial expansion  of terms in the parentheses }

(1/8) ( 3sqrt (3)  +  3[sqrt (3)]^2 i  + 3sqrt (3)i^2   + i^3)   =

(1/8)  ( 3sqrt (3)  + 9i   - 3sqr (3)  -  i)   =

(1/8)  ( 8i)   =      i

i  =   x

i^18   =   i^2  =    -1

i^12  =     1

i^6  =  -1

So

i^18   +   i^12  +  i^6   +  1   =

-1    +     1      -     1     +   1   =     0 

Wonderful Tiggsy!

Except, without a bloody diagram, I’m lost to where B’ and C’ are supposed to be, relative to the circle.  Are B’ and C’ positioned such that when they are joined by a vertical line, the line forms a tangent on the opposite side of the circle, mirroring the tangent of the line BC?

GA

Extend AB to B' and AC to C'.

Call angle CBO alpha, then angle OBB' = alpha, so angle ABC = 180 - 2(alpha).

Call angle BCO beta, then angle ACB = 180 - 2(beta).

From triangle ABC, (alpha + beta) = 115 deg, and then from triangle BOC, x = 65 deg.

(44.955 533 59, 301.421 852 9, 780.495 717 2, 1255.938 631, 1593.189 246, 1794.389 998, 1904.327 227, 1961.796 454, 1991.178 202, 2006.033 685, 2013.502 937, 2017.247 986, 2019.123 121, 2020.061 343, 2020.530 617, 2020.765 295, 2020.882 644, 2020.941 321, 2020.970 66, 2020.985 33, 2020.992 665, 2020.996 333, 2020.998 166, 2020.999 083, 2020.999 542, 2020.999 771, 2020.999 885, 2020.999 943, 2020.999 971, 2020.999 986, 2020.999 993, 2020.999 996, 2020.999 998, 2020.999 999, 2021, 2021, 2021, 2021, 2021 ==2021 

Slope of first line =   3

Slope of second line = 4/2  = 2

Slope of third line  =  9/3  =  3

Slope of fourth line =  (1/2)

Slope of fifth line     =    (1/4)

1 pair is good        one and  three