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25000 = 12500 e^(.09 t)     take ln both sides

ln25000 = ln12500  + (.09t)

(ln25000  -  ln12500 )  /  .09  = t    =  7.702 years

Solve for  x  and  y:                                x² - 11x - y  =  -24

                                                              x² - 11x + y  =  -24

Add down the columns:                       2x² - 22x       =  -48

Divide by 2:                                           x² - 11x        =  -24

Add 24 to both sides:                          x² - 11x + 24  =  0

Factor:                                                 (x - 8)(x - 3)  =  0

Solve:                                                 x  =  8     or     x  =  3

Replace  x  in the first equation:  x² - 11x - y  =  -24

                                               (8)² - 11(8) - y  =  -24               (3)² - 11(3) - y  =  -24

                                                     64 - 88 - y  =  -24                        9 - 33 - y  =  -24      

                                                           -24 - y  =  -24                            -24 - y  =  -24

                                                                  -y  =  0                                       -y  =  0

                                                                   y  =  0                                         y  =  0

                                                           --->     (8, 0)                                 --->   (3, 0)

You would have gotten the same  y-values had you replaced  x  in the second equation.

I think you may have left something out of your question.....

.00084 m = .084 cm

so      .084 x 2.99 = .25116  sq cm

x, y   =  2,-9    r = sqrt(1) = 1   

x,y   = 7, -4   r = sqrt(16) = 4   

Two whole numbers multiplied together won't result in a fraction....but you CAN convert it to a fraction

Example    5 x 3 =  15    x 3/3  =  45/3

                  6 x 4  = 24    x 2/2  =  48/2

Here is a graph from the desmos calculator:

6/3/8

1) 2/8

2)1/4 or .25

please give stars :)

2. If t is a real number, what is the maximum possible value of the expression -t^2 + 8t -4?

This is similar to (1).....this parabola turns downward.......so the t value that produces the max is:

-8 / [2(-1)] =   -8 / -2  =    4

Putting this value back into the function we have the max value of

-(4)^2 + 8(4) - 4 =   -16 + 32 - 4 =    12

1. What real value of t produces the smallest value of the quadratic t^2 -9t - 36

Since this is  an upwards-turning parabola, the "t" that produces the smallest possible value  is given by :

-b/ [2a] =  -[9] / [2 (1) ] =   9/2 = 4.5

I'm not too sure about this, but I don't believe any such numbers exist......here's my reasoning :

Let the 4 digits  be a , b, c and d    

And assume that each digit, except a, can only have a possible range of 0 - 9, inclusive  [a only ranges from 1 - 9, inclusive]

So we have  that

a + b + c + d   = 9    

Rearrange this as   (a + c) + ( b + d)  = 9      (1)

And it can be shown that if  the alternating sums of a 4 digit number equal some multiple of 11 (or, 0)...then that number is divisible by 11

For instance   1331.....we have  1 - 3 + 3 - 1  =  -2 + 2 =  0     and 1331  =  11^3

Or, for instance,  9141    .....we have   9 -1 + 4 - 1  =  8 + 3  = 11   and 9141  = 831 * 11

So.... our 4 digit number is divisible by 11  if    ( a - b) + (c - d ) = 11n    where n is an integer

Rearranging this, we have     (a + c) - (b + d)  = 11n        (2)

Adding (1)  and (2)   we have that

2(a + c)  =  11n + 9 

Now......the left side is even, so n must be odd...now, let n = 1.....then

2(a + c)  = 11(1) + 9

2 (a + c)  = 20

a + c  = 10

But this means that, by  (1),  10 + ( b + d ) = 9  → ( b + d) = -1.....but, since b, d can only range from 0-9, this is impossible

And we can clearly see that if n is any odd integer > 1, then a + c > 10  which by (1), makes (b + d) negative. And this violates our assumption that b, d can only range for 0-9, inclusive.

And n cannot be any negative odd, either....for instance, if n = -1

2(a + c)  = 11(-1) + 9

2(a + c)  = -2

a + c = -1    and this is also impossible, by our assumption

And for any negative odd > 1, (a + c) < -1 which also violates our assumption

Thus, no "n" exists that makes (1) and (2) both true, so there is no 4 digit number that will sum to 9 and also be divisible by 11

P.S. - I'd like some other mathematician(s) to look at this.......!!!!

0 hrs         500 mg

8 hrs         250

16             125

24               62.5

32               31.25

40               15.625

48 hrs           7.8125 mg

Set y = 61 and find x

61 = 14.7x + 16.9

x = (61 - 16.9)/14.7

x = 3

So spending will be more than $61 billion in all years after (2002 + 3), I.e. from 2006 onwards.

.

Use the calculator on the home page here.

In 3d with points (x1, y1, z1) and (x2, y2, z2) the distance formula is:

distance = sqrt( (x1-x2)^2 + (y1-y2)^2 + (z1-z2)^2 )

You need three coordinates for each point!

Total cost = d

Original cost per person = d/3

New cost per person = d/5

d/5 = d/3 - 11

11 = d/3 - d/5 

11 = 2d/15

11*15/2 = d

d = 82.5 dollars

Let 'x' = original cost to EACH of the three for gas....

then   3x = total gas cost

then two more join and the EACH COST goes down 11.00

so  x-11 is the new EACH COST

there are now 5 of them

5(x-11) = total cost = 3x 

5(x-11) = 3x

5x - 55 = 3x

2x = 55

x = 27.50    original cost of gas for 3    total cost = 3 (27.50) = 82.50   

WHen there are 5 the EACH COST is  82.50/5 =  16.50      which IS 11.00 than the original cost each.

30x = 42 (mod 47)

30x = 47n + 42.    n is an integer.

6(5x - 7) = 47n.   47 is prime so n must be a multiple of 6

n.         5x

1*6.     54.     Not divisible by 5

2*6.     101.   Not divisible by 5

3*6.     148.   Not divisible by 5

4*6.     195.  So x = 195/5.            

x = 39

There are calculators that will do this for you:                       

\(\sum_{1}^{22} x\) = 253

Basically it does this:  1 + 2 +3 + 4.....+ 22

The sum is given by:

S =   2 * [1 - 2^8 ]  / [ 1 - 2]  =   510

Find the standard form of the equation of the hyperbola with the given characteristics.

foci: (±4,0)      asymptotes: y= ±5x

We can answer this by process of elimination......

The center of the hyperbola is (0,0) and  the focus lies on the x axis so the hyperbola opens left/right.......so the form will be

x^2 / a^2   - y^2 / b^2  = 1

Then, obviously, answers c and  d are eliminated

The equation of one of the asymptotes will be  [ +b/a ] x  = 5x

Since  b/a =  5  then  [b/a]^2 = 25 →  b^2/a^2  = 25

But in answer "a,"  b^2 = 25 and a^2 = 16....then  b^2 / a^2 = 25/16 and this is < 25   so "a" is eliminated

And "e" is eliminated because a^2 > b^2....so b^2/a^2 < 1

So......answer "b" is correct

We have the following points on the parabola  (0,0), (200, 100), (-200, 100)

The vertex is the lowest point on the cable....this is located at (0, 0 )  and the parbola opens upward......so the form will be

y = ax^2

Since the point (200,100)  is on the parbola, we can sove for "a" thusly

100 = a(200)^2 

100 = a (40000)

a = 100/40000  =  1/400

So we have

y = (1/400)x^2      multiply both sides by 400

400y = x^2       →  " c "  is correct

The distance formula is    

d = √ [ (x₂ - x₁)^2 + (y₂ - y₁)^2 ]  

Let (6, -7) , (4, 4)  = (x₁,y₁)  (x₂, y₂)     so we have

d = √ [ (4 - 6)^2 + ( 4 - (-7) ) ^2 ]   =   √ [ (-2)^2 + (11)^2 ]  = √ [4 + 121 ]   = √ 125   =  √[25 * 5]  =  5√5 ≈  11.1803

(x−1)^2 +(y+8)^2 =16

Center=  ( 1, -8)   Radius =  4   →    'd"

(y-2)^2 +16(x-3)=0      put into this form

(x - h)   =  (1/[4p] ) (y -  k)^2       where (h,k) is the vertex  and the focus is given by [h + p , k]

(x - 3)  =  - (1/16)(y -2)^2

The vertex is (3, 2)       this parabola opens to the left, and we can find p thusly :

- [1/16 ] = [1/ (4p) =    so  p    =  - 4   

So the focus  = (3  + (- 4 ) ,  2 )  =  ( -1, 2)

"d"  is the correct choice

Expand the following:
(x/2+t)^4

(t+x/2)^4 = sum_(k=0)^4 binomial(4, k) (x/2)^(4-k) t^k = binomial(4, 0) (x/2)^4 t^0+binomial(4, 1) (x/2)^3 t^1+binomial(4, 2) (x/2)^2 t^2+binomial(4, 3) (x/2)^1 t^3+binomial(4, 4) (x/2)^0 t^4:
(binomial(4, 0) x^4)/16+1/8 binomial(4, 1) t x^3+1/4 binomial(4, 2) t^2 x^2+1/2 binomial(4, 3) t^3 x+binomial(4, 4) t^4

binomial(4, 0) = 1, binomial(4, 1) = 4, binomial(4, 2) = 6, binomial(4, 3) = 4 and binomial(4, 4) = 1:
t^4+(4 t^3 x)/2+6 t^2 (x/2)^2+4 t (x/2)^3+(x/2)^4

4/2 = (2×2)/2 = 2:
t^4+2 t^3 x+6 t^2 (x/2)^2+4 t (x/2)^3+(x/2)^4

Multiply each exponent in x/2 by 2:
t^4+2 t^3 x+6 t^2 (1/2)^2 x^2+4 t (x/2)^3+(x/2)^4

(1/2)^2 = 1^2/2^2:
t^4+2 t^3 x+1^2/2^2 6 t^2 x^2+4 t (x/2)^3+(x/2)^4

1^2 = 1:
t^4+2 t^3 x+(6 t^2 x^2)/2^2+4 t (x/2)^3+(x/2)^4

2^2 = 4:
t^4+2 t^3 x+(6 t^2 x^2)/4+4 t (x/2)^3+(x/2)^4

The gcd of 6 and 4 is 2, so (t^2 x^2×6)/4 = ((2×3) t^2 x^2)/(2×2) = 2/2×(3 t^2 x^2)/2 = (3 t^2 x^2)/2:
t^4+2 t^3 x+(3 t^2 x^2)/2+4 t (x/2)^3+(x/2)^4

Multiply each exponent in x/2 by 3:
t^4+2 t^3 x+(3 t^2 x^2)/2+4 t (1/2)^3 x^3+(x/2)^4

(1/2)^3 = 1^3/2^3:
t^4+2 t^3 x+(3 t^2 x^2)/2+1^3/2^3 4 t x^3+(x/2)^4

1^3 = 1:
t^4+2 t^3 x+(3 t^2 x^2)/2+(4 t x^3)/2^3+(x/2)^4

2^3 = 2×2^2:
t^4+2 t^3 x+(3 t^2 x^2)/2+(4 t x^3)/2×2^2+(x/2)^4

2^2 = 4:
t^4+2 t^3 x+(3 t^2 x^2)/2+(4 t x^3)/(2×4)+(x/2)^4

2×4 = 8:
t^4+2 t^3 x+(3 t^2 x^2)/2+(4 t x^3)/8+(x/2)^4

4/8 = 4/(4×2) = 1/2:
t^4+2 t^3 x+(3 t^2 x^2)/2+(t x^3)/2+(x/2)^4

Multiply each exponent in x/2 by 4:
t^4+2 t^3 x+(3 t^2 x^2)/2+(t x^3)/2+(1/2)^4 x^4

(1/2)^4 = 1^4/2^4:
t^4+2 t^3 x+(3 t^2 x^2)/2+(t x^3)/2+1^4/2^4 x^4

1^4 = 1:
t^4+2 t^3 x+(3 t^2 x^2)/2+(t x^3)/2+x^4/2^4

2^4 = (2^2)^2:
t^4+2 t^3 x+(3 t^2 x^2)/2+(t x^3)/2+x^4/(2^2)^2

2^2 = 4:
t^4+2 t^3 x+(3 t^2 x^2)/2+(t x^3)/2+x^4/4^2

4^2 = 16:
Answer: |t^4 + 2t^3 x + (3t^2x^2)/2 +( tx^3)/2 + x^4/16