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It is 972972

In Tina's Stamp collection, 70% of the stamps are Canadian and the rest are International. Tina has 500 more Canadian stamps than International stamps.How many stamps does she have.

Let Canadian stamps=C

Let International stamps=N

C=875     and        N=375

C - 500=N,  .70(N + C) =C

Solve the following system:
{C-500 = N |     (equation 1)
0.7 (C+N) = C |     (equation 2)
Express the system in standard form:
{-N+C = 500 |     (equation 1)
0.7 N-0.3 C = 0 |     (equation 2)
Add 0.7 × (equation 1) to equation 2:
{-N+C = 500 |     (equation 1)
0 N+0.4 C = 350 |     (equation 2)
Divide equation 2 by 0.4:
{-N+C = 500 |     (equation 1)
0. N+C = 875. |     (equation 2)
Subtract equation 2 from equation 1:
{-N+0 C = -375. |     (equation 1)
0 N+C = 875. |     (equation 2)
Multiply equation 1 by -1:
{N+0 C = 375. |     (equation 1)
0 N+C = 875. |     (equation 2)
Collect results:
Answer: |  N = 375        and             C=875
 

Take the integral:
 integral cos(x) sinh(x) dx
For the integrand cos(x) sinh(x), integrate by parts,  integral f dg = f g- integral g df, where
 f = cos(x),     dg = sinh(x)  dx,
 df = -sin(x)  dx,     g = cosh(x):
 integral cos(x) sinh(x) dx = cos(x) cosh(x)+ integral sin(x) cosh(x) dx
For the integrand sin(x) cosh(x), integrate by parts,  integral f dg = f g- integral g df, where
 f = sin(x),     dg = cosh(x)  dx,
 df = cos(x)  dx,     g = sinh(x):
 integral cos(x) sinh(x) dx = sin(x) sinh(x)+cos(x) cosh(x)- integral cos(x) sinh(x) dx
Add  integral cos(x) sinh(x) dx to both sides:
2 integral cos(x) sinh(x) dx = sin(x) sinh(x)+cos(x) cosh(x)+constant
Divide both sides by 2:
Answer: |   = 1/2 (sin(x) sinh(x)+cos(x) cosh(x))+constant

Let's call the first number, A   ....    and it can be represented as  10m + 7

And the second number can be B.......and it can be represented as 10n + 5

So...the sum of these numbers =  10m + 10 n + 7 + 5 = 10m + 10n + 12   =  10(m +n) + 10 + 2   =  10(m + n + 1) + 2

And 9 times this =  90(m + n + 1) + 18  =  90(m + n + 1) + 10 + 8

And dividing this number by 10  is equivalent to dividing each separate term by 10  =

9(m + n + 1)  + 1   +  8/10

q + 8/10

The sum of the first two terms produce some integer, q, so.......the remainder is 8.....and Solveit is correct  !!!!!

We have five constraints....two of them,  x ≥ 0  and y ≥ 0,   just indicate that our answer - if any - will be found in the first quadrant

Here's a graph of the other three : https://www.desmos.com/calculator/tvty6v1szo

It can be shown that the objective function  z = 300x + 150   is maximized at some corner point(s) where these three inequalities intersect......here, we only have one corner point intersection....and  that occurs at (120, 90)

So.....z_max =  300(120) + 150  =  36,150

2*4*4*8*5*6*2*4*1*4*5*36*5*4*4/4*1*8*4*2*51*556159/1 = 1606063893774336000

1606063893774336000 = 1.606063893774336e18

Part A   = 29

Part B  = 4

Part C   =     17 out of 50 were in the band.......this is 34%........and applying this percentage to 250  =  250 * .34  = 85 students expected to participate in the first practice

It looks like you have complted the table

The answer to the first question comes from the intersection of the second row with the second column  = 5

The answer to the second =

N(Selfie or Text Message)  = N(Selfie) + N(Text Message) - N(Selfie and Text Message)  =

                                                 38          +      69           -                22     =

                                                        107             -       22

                                                              85

So the percent that have taken a selfie or sent a text message divided by the total surveyed =

85 / 90    =  17/18   = about   94%

D is correct

The probability of selecting a purple tulip form the second vase on the second draw depends upon what happened in the first draw

Example :

P(Yellow on first Draw and Purple on the Second Draw)  = 9/14 * 5/13  =  45 / 182 

But

P (Purple on first Draw and Purple on the Second Draw)  = 5/14 * 4/13  = 20 / 182  

Thus....we have a  greater chance of drawing Yellow-Purple on the first two draws than of drawing Purple-Purple on the frist two draws

We have this form :  y = ax^2 + bx + c    .....so......

y  = x^2 - 6x + 7      

The x coordinate of the vertex is given by   -b/ 2a      where b = -6  and a   = 1

So   .....    - (-6)/ 2(1)  =  6/2  = 3    and this is the x coordinate of the vertex

To find the y coordinate......just plug "3" back into the function....and we have

(3)^2  - 6(3) + 7   =  

9 - 18 + 7  =

-2

So  the vetex  is  (3, -2)

Here's the graph that confirms this :  https://www.desmos.com/calculator/2h7phyucv7

For two events to be independent, we must have

P(A) * P(B)  = P(A and B)

Hence

P(Passing Math) * P(Passing History)  = P(Passing Math and History)     ????

(.80)                   *        (.70)          =        (.56)

And this is true, so.......

The events are independent   and  "C"  is correct

OK......thanks, Alan  !!!!!

5      9      14

24    2      26

29   11     40

N (Algebra or Physics)   = N(Algebra) + N(Physics) - N(Algebra and Physics)   =

                                            29             +     14          -        5                    =

                                                   43     -    5    = 

                                                        38

15   40    55

55   10    65

70   50   120

Tablet and no smartphone   =  40

So    40 / 120   = 1/3    =  33 1/3 %

And you are correct    !!!!

1=

Take the integral:
 integral sqrt(2-x) x^2 dx
For the integrand sqrt(2-x) x^2, substitute u = sqrt(2-x) and  du = -1/(2 sqrt(2-x))  dx:
  =  -2 integral u^2 (2-u^2)^2 du
Expanding the integrand u^2 (2-u^2)^2 gives u^6-4 u^4+4 u^2:
  =  -2 integral (u^6-4 u^4+4 u^2) du
Integrate the sum term by term and factor out constants:
  =  -2 integral u^6  du+8 integral u^4  du-8 integral u^2  du
The integral of u^6 is u^7/7:
  =  -(2 u^7)/7+8 integral u^4  du-8 integral u^2  du
The integral of u^4 is u^5/5:
  =  (8 u^5)/5-(2 u^7)/7-8 integral u^2  du
The integral of u^2 is u^3/3:
  =  -(2 u^7)/7+(8 u^5)/5-(8 u^3)/3+constant
Substitute back for u = sqrt(2-x):
  =  -2/7 (2-x)^(7/2)+8/5 (2-x)^(5/2)-8/3 (2-x)^(3/2)+constant
Which is equal to:
Answer: |  =  -2/105 (2-x)^(3/2) (15 x^2+24 x+32)+constant

2-

integral x sec^2(x) 2 x dx = 2 (x (-i x+2 log(1+e^(2 i x))+x tan(x))-i Li_2(-e^(2 i x)))+constant

3-

Take the integral:
 integral x/sqrt(x+5) dx
For the integrand x/sqrt(x+5), substitute u = x+5 and  du =   dx:
  =   integral (u-5)/sqrt(u) du
Expanding the integrand (u-5)/sqrt(u) gives sqrt(u)-5/sqrt(u):
  =   integral (sqrt(u)-5/sqrt(u)) du
Integrate the sum term by term and factor out constants:
  =   integral sqrt(u) du-5 integral 1/sqrt(u) du
The integral of sqrt(u) is (2 u^(3/2))/3:
  =  (2 u^(3/2))/3-5 integral 1/sqrt(u) du
The integral of 1/sqrt(u) is 2 sqrt(u):
  =  (2 u^(3/2))/3-10 sqrt(u)+constant
Substitute back for u = x+5:
  =  2/3 (x+5)^(3/2)-10 sqrt(x+5)+constant
Which is equal to:
Answer: |  =  2/3 (x-10) sqrt(x+5)+constant

i think:

x is a multiple of 10

\(a=x+7 \\ b=x+5 \\ (a+b)\times9=(2x+12)\times9\\ \frac{18x+108}{10}=\frac{(18x+100)+8}{10} \)

so the remainder is always 8

You add the remainder of 7 +5=12 . No matter what 2 numbers you choose, the number will always end in 2. And when you multiply this by 9, the number will always end up in 8. Hence the reainder will always be 8.

Eamaple: 17/10=remainder is 7. 15/10=remainder is 5. Whether you add the numbers we chose:17+15=32 or their reainders: 7+5=12.

ok I stand corrected.  Thanks :)

MELODY:

25! ends in 6 trailing zeros.155112 1004333098 5984,000,000

25/5=5 and 25/25=1

5  +  1  =6 zeros

 \(\begin{array}{rcll} 4x^2 &=& 3x\\ 4x^2-3x &= & 0\\ x(4x-3) &=&0\\ \boxed{ x_1=0\;\;\; x_2=\frac34} \end{array}\)

Here is a general rule to find number of trailing zeros in a factorial:

DIVIDE THE NUMBER BY POWERS OF 5.

Example: 100!

100/5^1 +100/5^2 +100/5^3.........etc

20           +  4             +    0      =24 zeros in 100 factorial. Disregard any fractions. Just take the integer part.

935*0.6 = 561

a) 10!

1*2*3*4*5*6*7*8*9*10

1*2*3*4*5*6*7*8*9

one

1*3*4*6*7*8*9

two

1*3 ends in 3

3*4 ends in 2

2*6 ends in 2

2*7 ends in 4

4*8 ends in 2

2*9ends in 8

so the last 3 digits are 800

that is 2 zeros at the end.

b) 14!

10!  ends in 800

11! ends in 800

12! ends in 600

13! ends in 800

14! ends in 200

c) 20!    

15! ends in 000

20! ends in 0000        4 zeros

d) 25!

1*2*3*4*5 has 0ne zero at the end

so  add another zero

25!                           5 zeros