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distance / rate = time

upriver    11 / (r-c)  = 4.4 hr         r = speed if paddling in still water         c = current speed

             re-arrange to     c = (11-2r)/2

    downriver  11/ (r+c) = 2     sub in the red equation then sovle for r

                    r = 4 km/hr     

Use this result in either of the upstream or downstream equations to find c = 1.5 km/hr

If no current    :   11 km / 4 km/hr = 2.75 hrs       2 hrs 45 min

Solve the inequality x^3 + 4x > 5x.

x^3-x>0

x(x+2)(x-1)>0

-1x

x>1

Yay!

x = money at 5% interest

10000 -x    = money at 16%

.05 x  +  .16 ( 10000 - x ) = 775

x = 7500 at 5%        then 2500 at 16 %

Their differnce in speed is   2 mph

1/6 hr * 18 m/hr = 3 miles to go when Anna finishes

Anna gets ahead 2 miles every hour          3 miles /2m/hr = 1.5 hr for anna to finish       1.5 hr x 20 m/hr = 30 mile bike race

Take 1/2 of the x coefficient ....then square it

5/4 * 1/2 = 5/8         square it:   25 / 64

x^2 + 5/4 x + 25 / 64                is equal to   (x + 5/8)²

Do you mean 2^3√4 or 2^(3√4)?

w r o n g : / 

pretty sure it reads: Suppose a and b are different prime numbers greater than 2. How many whole-number divisors are there for the integer a(2a + b) - 2a^2 + ab

How about writing it so it is more legible.  

Thanks EP

Are you happy with the answer nimi?   

I see you have given EP a point, that is nice, but It would be good if you responded as well.

It is good to see you here Geno  

the gradient of the line joining those points is    (-2- -7)/(3- -1) = 5 / 4 = 1.25

so the gradient from either of those points to (-6,y) must also be  1.25

\(\frac{y- -2}{-6 - 3}=1.25\\ \frac{y+2}{-9}=1.25\\ y+2=-11.25\\ y=-13.25\)

You can graph those points to check that they are collinear  ( in a line)

I didn't get a single result I got this graph

I worked on the basis that x cannot be 3

and if 

|x|>3 then  x+3y+x^2-9 = 0

and if

|x|<3 then  x+3x-x^2+9=0

I do not understand what this question is asking.

Maybe another mathematician would alkie to comment ?

I graphed it

then I added X(1,1)  and E(5,1)

Area = area of triangle ABX +  Area of trapezium XBCE  -  area of triangle DCE

Area = 1+ (3/2 * 4) - 0.5

Area = 1 + 6 - 0.5

Ara = 6.5 unites square

(Sorry if I ended up posting this twice, my chromebook glitched)

I ran the numbers, and got

four hundred twenty octillion five hundred fifty-three septillion four hundred forty-nine sextillion three hundred quintillion one hundred twenty quadrillion two hundred seven trillion six hundred billion five hundred twenty-six million five hundred five thousand eight hundred forty-five

PLEASE DONT EVER USE THIS EXPANSION, NOBODY WILL UNDERSTAND YOU (instead use scientific notation, or even standard form).

I'm pretty sure it uses the short scale of numbers, as opposed to the long scale.

https://www.calculatorsoup.com/calculators/conversions/numberstowords.php

Answer: \(6.5\) \(units²\)

Solution:

I would think that the best way is to graph the shape (although there is probably some better way) and then to break it up into three triangles and a rectangle. You can find the area of a triangle by looking at the graph, or just by looking at its points. (1-0)(3-1)/2 = 1 for example on the first triangle. I'm not great at explaining, so sorry if this doesn't make sense.

Triangle 1: (0,1), (1,1), (1,3)

The area of Triangle 1 is 1 square unit.

Triangle 2: (1,3), (1,2), (5,2)

The area of Triangle 2 is 2 square units.

Triangle 3: (5,2), (4,2), (4,1)

The area of Triangle 3 is 0.5 square units.

Rectangle: (1,1), (1,2), (4,2), (4,1)

The area of the rectangle is 3 square units.

Adding them all up you get 1 + 2 + 0.5 + 3 = \(6.5\) \(units²\)

answered.

19 pencils is 3 boxes of 7pencils   (you have to pay for the extra 2 even though you do not want them)

3 pens is 1 boxes of 5 pens   (you have to pay for the extra 2 even though you do not want them)

So Roy needs to buy 3 boxes of pencils and 1 box of pens.  How much is that going to be?

420 553 449 300 120 207 600 526 505 845

With long lumbers like this you would never given them a name, I mean not like you are referring to.

Anyone who actually deals with large numbers like this would refer to it as

4.20553449300120207600526505845 * 10^29

This is written in scientific notation [which can also be referred to as standard form (edited, form is the right word, thanks Why)  if you are in the UK.]

Scientific notation is the better name though as it will be understood throughout the English speaking world.

Show the working.  Steps 1, 2, 3 and 4      

When the positive integer x is divided by each of 4, 5, 6, and 7,

it has a remainder of 3.
What is the sum of the three smallest possible values of x?

\(\begin{array}{|rcll|} \hline x&\equiv& 3 \pmod{4} \\ x&\equiv& 3 \pmod{5} \\ x&\equiv& 3 \pmod{6} \\ x&\equiv& 3 \pmod{7} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{sum} &=& \Big( 3+0\times \text{lcm}(4,~5,~6,~7) \Big) \\ && +\Big( 3+1\times \text{lcm}(4,~5,~6,~7) \Big) \\ && +\Big( 3+2\times \text{lcm}(4,~5,~6,~7) \Big) \\\\ \text{sum} &=& 3 + ( 3+420 )+( 3+2\times420 ) \\\\ \text{sum} &=& 3 + 423+843 \\ \text{sum} &=& 1269 \\ \hline \end{array}\)

What is the largest number that will always divide the sum of any 8 consecutive positive integers?

My attempt:

\(\begin{array}{|rcll|} \hline \text{sum} &=& \dfrac{(a_1+a_8)}{2}\times 8 \\\\ \text{sum} &=& (a_1+a_8)\times \color{red}4 \\ \hline \end{array} \)

The largest number is \(\color{red} 4\)

Find length BC.

\(\text{Let $BC = \color{red}x$}\)

\(\begin{array}{|lrcll|} \hline (1): & AE^2+DE^2 &=& 4^2 \\ (2): & DE^2+CE^2 &=& 6^2 \\ (3): & AE^2+BE^2 &=& 3^2 \\ (4): & BE^2+CE^2 &=& x^2 \\ \hline \end{array}\)

\(\small{ \begin{array}{|lrcll|} \hline (3)-(1)+(2): & (AE^2+BE^2)-(AE^2+DE^2)+ (DE^2+CE^2) &=& 3^2-4^2+6^2 \\ & BE^2+CE^2 &=& 3^2-4^2+6^2 \\ & BE^2+CE^2 &=& 29 \quad | \quad BE^2+CE^2 = x^2 \quad (4) \\ & x^2 &=& 29 \\ & \mathbf{x} &=& \mathbf{\sqrt{29}} \\ \hline \end{array} }\)

\(BC = \sqrt{29}\)

Let angle BAD = 90º

BD = 5

AX = 2.4

BX = 1.8

DX = 3.2

CX = sqrt(CD² - DX²) 

BC = sqrt(BX² + CX²) = √29