CPhill

1. Find the second smallest positive integer that gives a remainder of 2 when divided by 3 and gives a remainder of 3 when divided by 7.

We have that

N = 3a + 2

N = 7b + 3

Subtracting these equations we have that

3a - 7b - 1  = 0

3a - 7b =  1

And the second smallest positive integer results when a = 12 and b = 5

So.....the second smallest integer is 38.....just as Max said !!!!

This is very similar to the one I did here:

http://web2.0calc.com/questions/can-someone-please-help-me-with-this-question-and-the-one-i-just-posted-i-m-very-stuck-and-need-to-finish-these-two-questions-to-complete-my-class-please

See if you can follow my example to work this one....if you get stuck, let me know

Find a set of parametric equations for the line that passes through the given points.

(3, 7/2, -3/2), (9/2, 3/2, 1)

The vector connecting these points is given  by  < 9/2 - 3, 3/2 - 7/2, 1 - (-3/2) .  =  < 3/2, -2, 5/2 >

So...using either of the given points [ I'll use the first one ].....the vector form of the line is :

< 3, 7/2, -3/2 > + t <3/2, -2, 5/2>   =  < 3 + (3/2)t, 7/2 - 2t, (-3/2) + (5/2)t >

And the parametric equations are

x = 3 + (3/2)t

y = 7/2 - 2t

z = (-3/2) + (5/2)t

Verify that (9/2, 3/2, 1)   lies on this line when t = 1

Look at the graph, here:  https://www.desmos.com/calculator/rouxa8yruw

As x approaches both  0 and 9 from the right, y approaxhes infinity

(-3*15)+13^2+2

(-45) + 169 + 2 = 

126

5 / (z^2 - 8z + 7 )  =  1 / [4z + 4]   cross-multiply

5 [ 4z + 4]  = z^2 - 8z + 7   simplify

20z + 20  = z^2 - 8z + 7       subtract 20z + 20 from both sides

0  = z^2 - 28z - 13

Using the quadratic formula....the solutions  are    14 ± √209

Note, Max, that by inspection, at least one real solution  is {x,y}  = {1, 1}

If a conical tank with a radius of 8 feet, and a height of 12 feet is being filled with water a rate of 2 cubic feet of water per minute, find how fast the water level is rising when the depth of the water in the tank is 2 feet. 

Notice that the ratio of the height to the radius = 12/8  = 3/2

So....at any time, t,    r = (2/3)h

And we can  represent the volume as :

V =  (1/3)pi * r^2  * h  =   (1/3) [ (2/3)h] ^2 *h =  (1/3)pi (4/9)h^3  =  (4/27)pi*h^3

Take the derivative with respect to time, t

dv/dt  = 3*(4/27)pi*h^2* dh/dt

dv/dt = (12/27)pi* h^2 * dh/dt

dv/dt  = (4/9)pi*h^2* dh/dt

And dv/dt = 2ft^3/min.....and we need to solve for dh/dt  when h = 2ft

So we have

2ft^3/min  = (4/9)pi (2ft)^2 *  dh/dt

2ft^3/min  = pi*(16/9)ft^2 * dh/dt

[2] / [pi*(16/9)] ft/min = dh/dt  = 9 / [8 pi] ft/min  ≈ 0.358 ft/min

Just enter the fractions........rhe calculator can handle all four basic operations

√ [ 2 + 3√x - x]  + √ [6 - 2√x - 3x ]   = √10 + 4√x - 5x]

Note, Max......that, by inspection, x = 1   is at least one solution.......let's work through this

Square both sides

[ 2 + 3√x - x]  + 2√ [ 2 + 3√x - x] * √ [6 - 2√x - 3x ] + [6 - 2√x - 3x ]   = 10 + 4√x - 5x

Simplify

8 + √x - 4x  + 2 √ [( 2 + 3√x - x) (6 - 2√x - 3x) ] =   10 + 4√x - 5x

2 √ [( 2 + 3√x - x) (6 - 2√x - 3x) ]  =  2 + 3√x -x

2 √  [-7 x^(3/2)+3 x^2-18 x+14 sqrt(x)+12 ]   = 2 + 3√x -x

Square both sides again

4 [-7 x^(3/2)+3 x^2-18 x+14 sqrt(x)+12 ]  = -6 x^(3/2)+x^2+5 x+12 sqrt(x)+4

-28x^(3/2) + 12x^2 - 72x + 56sqrt(x) + 48  =  -6 x^(3/2)+x^2+5 x+12 sqrt(x)+4

22x^(3/2) - 11x^2 + 77x - 44sqrt(x) - 44  =  0

Divide through by 11

2x^(3/2) - x^2 + 7x - 4sqrt(x) - 4  = 0

This is a little difficult to factor......WolframAlpha gives

-(sqrt(x)-1) (sqrt(x)+2) (x-3 sqrt(x)-2) = 0

Setting the first factor to 0  →  sqrt(x) - 1  = 0       and it's obvious that  x = 1

Setting the second factor to 0  → sqrt (x) + 2  = 0  →  sqrt (x)  = -2    and this is a non-real result

Setting the third factor to 0  →  x - 3sqrt(x) - 2  = 0

And WolframAlpha  gives  [ 13 + 3√17 ] / 2    as the other real solution

So....the real solutions are  x = 1   and x = [ 13 + 3√17 ] / 2