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1  63/100    

x = bike time      6-x = walk time

44x    +   4 (6-x)    = 164         solve for x          (He bikes at 44 mph!   Wow! )

Use distance formula to find the distance between the two points....this is the side length of the square....then area = side X side

d^2 = ( -1 -2)^2 + ( 4 - -8)^2        solve for d  ( the side length)

2240

x = original amount

x - 1540       amount after sofa purchase

   spending 1/5   means 4/5 is left

       4/5 ( x - 1540)       and this equals    1/4x

4/5 (x - 1540) = 1/4 x               Now, just solve for 'x'  ......

sqrt(2009*2010*2011*2012+1)==4042109

A jar contains n nickels and d dimes. There are 22 coins in the jar, and the total value of the coins is $1.60. How many nickels and how many dimes are in the jar?

n  +  d ==22........................................(1)

0.05n  +  0.10d ==1.60......................(2), solve for d, n

Use substitution to get:

d==10 dimes    and  n==12 nickels.

Let the number of apples be x

then

x - x/5 - 80 = 3x/10

4x/5 - 3x/10 = 80

5x = 80 * 10

x = 160

------

then on monday he sold 32 apples

tuesday = 80 apples

then he should have 48 apples remaining

so

3*160/100 = 48

so the answer is 160 apples

For each positive integer n, let S(n) denote the sum of the digits of n. How many three-digit n's are there such that S(n) = n (mod 9) ?

I think this last bit: S(9)==n (mod 9) should read: n==S(n) mod 9, which means n mod 9==sum of digits of n.

Example: n=313 and 313 mod 9 ==7 which is the sum of digits of 313.

If that is correct, then you have 120 such n as follows:

(100, 101, 102, 103, 104, 105, 106, 107, 110, 111, 112, 113, 114, 115, 116, 120, 121, 122, 123, 124, 125, 130, 131, 132, 133, 134, 140, 141, 142, 143, 150, 151, 152, 160, 161, 170, 200, 201, 202, 203, 204, 205, 206, 210, 211, 212, 213, 214, 215, 220, 221, 222, 223, 224, 230, 231, 232, 233, 240, 241, 242, 250, 251, 260, 300, 301, 302, 303, 304, 305, 310, 311, 312, 313, 314, 320, 321, 322, 323, 330, 331, 332, 340, 341, 350, 400, 401, 402, 403, 404, 410, 411, 412, 413, 420, 421, 422, 430, 431, 440, 500, 501, 502, 503, 510, 511, 512, 520, 521, 530, 600, 601, 602, 610, 611, 620, 700, 701, 710, 800) >>Total No = 120 such poitive integers.

AE= sqrt(17^2 + 30^2) = 

\(AE=\sqrt{17^2+30^2}\\ AE= \sqrt{1189}\;cm\\~\\ EC=\sqrt{13^2+13^2}\\ EC=\sqrt{338}\;cm\\~\\ AC=\sqrt{17^2+4^2}\\ AC=\sqrt{305}\\ \)

Heron's formula for finding area.

Let a,b,c   be the lengths of the sides of a triangle. The area is given by:

\(area=\sqrt{p(p-a)(p-b)(p-c)}\qquad \text{where p=half the perimeter}\\~\\ p=\frac{\sqrt{1189}+\sqrt{338}+\sqrt{305}}{2}\\~\\ \)

You can do the substitutions to find the answer. 

If f(x) and g(x) are polynomials such that f(x) + g(x) = -2 + x, then what is g(x) if f(x) = x^8 - 2x^3 + 2?

          f(x)            + g(x) = -2 + x

  x^8 - 2x^3 + 2   + g(x) = -2 + x

 g(x) =  - x^8 +  2x^3  - 2 + x - 2

 g(x) =  - x^8 +  2x^3 + x - 4

Since 1 + 10 + 27 (the coefficients of the x^3, x^2 and x terms respectively) make 38, the root is x = 1.

Hey guest,

Try forming cases, if n == {0, 1, 2, ... ,8} mod 9

x = 85/3.

ADC =114 so we know BDC is then equal to 66 (180-114)

DBC is also equal to BDC because its an isosceles traingle, DBC and BDC = 66

180-(66+66)=48 

because AB and AC are equal sides, ABC and ACB are both equal to 66

66-48=12 :)

240 / 60 = 40 + 30 = 70 + 10 = 80

The site did something weird with my dollar signs.  

Those two lines should read  

It can't be 30 because that would mean there are no items at 2 dollars or 3 dollars       

It can't be 20 because that would leave only 10 items at 2 dollars or 3 dollars and that isn't enough   

_.

Janice bought 30 items each priced at 30 cents, 2 dollars, or 3 dollars. If her total purchase price was $48.00, how many 30-cent items did she purchase?      

The total purchase price is an even number of dollars  

Therefore the number of 30¢ items has to be a multiple of 10  

It can't be 30 because that would mean there are no items at $2 or $3    

It can't be 20 because that would leave only 10 items at $2 or $3 and that isn't enough   

Conclusion:   There must be 10 items priced at 30¢  

_.

On the graph of y = (x + 2)^2 - 10, how many points are there whose coordinates are both negative integers?  

infinite?  I don't understand what you want, so I suggest you copy that equation and

paste it in to https://www.desmos.com/calculator and count the points you want.  

Let no of apples in carton = x

When Packed in days of 4. then remaining

apples 3.

-------------------------------------------------------------------------------------------

When Packed in days of 6 then Remaining

apples = 3.

for finding least number we will find the

LCM of two numbers . (4,6)

       4 = 2 * 2 * 1

       6 = 2 * 3 * 1 
   LCM= 2 * 2 * 3 * 1 = 12

   (4,6)

Now remainder is always 3. So add 3.

in LCM. (12 + 3) = 15.

[ Total apples  =  15 apples. in carton. ]

Let's show our steps by getting an answer instead of saying an answer without showing steps and also not make mistakes.

Number of non-venomous violet vipers

                                                 = 23

 Number of non-venomous vermillion

                 vipers = 23 - 10 = 13

 Total number of non-venomous

      vipers = 23 + 13 = 36

Number of Venomous vipers

           = 36 [ Venoumous vipers

                      = Non-Venomous vipers ]
Number of venomous violet vipers

                      = 3/4 * 36 = 27

So, Number of venomous vermillion

                  Vipers = 1/4 * 36 = 9

      There are \(\boxed{9}\)venomous vermilion vipers in Vennsylania zoo.

How many sweets does sandra have in the container in the end?

Hello Guest!

\(\color{blue}z+p+10=\frac{2}{3}(z+p+10)+z\\ p+10=\frac{2}{3}z+\frac{2}{3}p+\frac{20}{3}\\ \frac{p}{3}=\frac{2z}{3}+\frac{20}{3}\)

\(p=2z+20\)

\( \color{blue}z+p+10+30=\frac{3}{4}(z+p+40)+z\\ p+40=\frac{3}{4}z+\frac{3}{4}p+30\\ \frac{p}{4}=\frac{3}{4}z-10\)

\(p=3z-40 \)

\(2z+20=3z-40\)

\(z=60\)

\(p=2z+20\)

\(p=140\)

\(140+10+30+60=\color{blue}240\ sweets\\ does\ Sandra\ have\ in\ the\ container\ in\ the\ end.\)

  !

\cfrac{1}{1+ \cfrac{1}{\sqrt{3}+2}}.

\(\cfrac{1}{1+ \cfrac{1}{\sqrt{3}+2}}\\ 1\div\left[ 1+\cfrac{1}{\sqrt{3}+2} \right]\\ 1\div\left[ \cfrac{\sqrt3+2+1}{\sqrt{3}+2} \right]\\ 1\div\left[ \cfrac{\sqrt3+3}{\sqrt{3}+2} \right]\\ 1*\left[ \cfrac{\sqrt3+2}{\sqrt{3}+3} \right]\\ \cfrac{\sqrt3+2}{\sqrt{3}+3} * \cfrac{\sqrt3-3}{\sqrt{3}-3} \\ \)

You can check what I have done and take it from there.