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(i)      -5t^2  +  15t  +  1   =  11

          -5t^2  +  15t  - 10   =  0       divide through by -5

            t^2   -  3t   +  2     =  0       factor

            (t  - 2)  ( t - 1)   =   0

             At 11m  at     1sec    and  2 sec

(ii)       -5t^2  + 15t  +  1   =  0

             5t^2   - 15t  -  1   =  0

            Hits  the  ground  at  ≈  3.06 sec     (from WolframAlpha)

(iii)     Greatest  height reached  at time    (-15) / (2 (-5) )   =   15/10  =  1.5 sec

Max  ht =    -5(1.5)^2  + 15 (1.5)   + 1   =   12.25  m

Call  the edges  x   , rx  and r^2  x

So

x ( rx ) ( r^2 x)  =  216   ⇒   x^3 r^3  =  216   ⇒   xr  =  6   ⇒   r = 6/x

And

2 ( x * xr   +  x * xr^2  + xr * xr^2)     =   516

x^2r  + x^2r^2  + x^2r^3   =  258

x^2 (6/x) + x^2 ( 6/x)^2 +  x^2 ( 6/x)^3  =   258

6x   +  36  +  216/x =   258          divide through by 6

x +  6  +  36/x  =  43                 mutiply through by x

x^2  +6x  + 36  =  43x

x^2 - 37x  + 36 =  0

(x - 1) ( x - 36)  =  0

x =  1   or x  =  36

Take  the first  value  and  r =  6 /1   =   6

The longest  edge  is    1^2 * 6^2  =   36

3a  +  4b    =   7   ⇒  multiply through by 3 ⇒    9a  + 12b  = 21

6a  +  4b  =  k  - 3a   - 8b ⇒ simplify ⇒               9a + 12b  =  k

k = 21

                           Music              Not Music

Left handed         32                       8                 40

Right handed       78                     32               110

                           110                     40              150

P( Right Handed l Taking Music)  =   78/ 110   =    39 / 55   ≈    71%

Looks to be missing something.....like.....everything  ......

(5-2)^3  =  3^3   = 27   do not have  any paint on them

consider Sum of money shared by both =x

Total: 3+4=7

Jonathan shared money = 3*x/7

Kenneth shared money = 4*x/7

Given

3x/7 -1/6*(3x/7) = 4x/7-15

3x/7-x/14 = 4x/7-15

5x/14 = 4x/7 -15

5x/14-4x/7=-15

-3x/14=-15

3x=15*14

x=5*14

x=70 $

Jonathan shared money = 3/7*70=30$

Kenneth shared money = 4/7*70=40 $

3 x 3 x 3 with NO paint

3 x 3 x 6 with ONE face painted   ....sorry.... didn't read the Q entirely

The length of the longest edge is 28.

Nevermind Incorrect

Let's solve this problem by working backwards. 
 

Third Round:

Since Farhan gave half of his marbles last round and he now has 25 marbles, he had 25 x 2 = 50 marbles in the third round.

Darius has 45 - 25 = 20 marbles in the third round as well.

Second Round:

Ellen gave a fourth of her marbles during the second round and now has 30 marbles left. This means that 3/4 of her previous marbles = 30, so she has 30 x 4/3 = 40 marbles in the second round.

Ellen also gave 10 marbles to Farhan during this round, so Farhan had 50 - 10 = 40 before then.

First Round:

Darius gave 1/5 of his marbles and ended up with 20 marbles. This means that Darius originally had 20 x 5/4 = 25 marbles.

Ellen received 5 marbles and ended up with 40 marbles, so she originally had 40 - 5 = 35 marbles. 

Let's now write down the number of marbles each person has after working backwards:

Ellen has 35 marbles.

Darius has 25 marbles.

Farhan has 40 marbles.

Correct!

Incorrect

Information given:

At the end,

Number of marbles Darius has = 45 marbles

Number of marbles Ellen has = 30 marbles

Number of marbles Farhan has = 25 marbles

Calculations: Let’s say Darius has x number of marbles, Ellen has y number of marbles, and Farhan has z number of marbles at the beginning.

1st round

Darius gave 1/5 marbles to Ellen.

=> Number of marbles Darius has = x – x/5 = 4x/5

Number of marbles Ellen has = y + x/5

2nd round

Ellen gave 1/4 marbles to Farhan

=> Number of marbles Ellen has = (y + x/5) – 1/4 (y + x/5) = 3/4 (y + x/5)

Number of marbles Farhan has = z + 1/4 (y + x/5)

3rd round 

Farhan gave 1/2 of his marbles to Darius.

=> Number of marbles Farhan has = 1/2 [z + 1/4 (y + x/5)]

Number of marbles Darius has = 4x/5 + 1/2 [z + 1/4 (y + x/5)]

So, it can be written as,

1/2 [z + 1/4 (y + x/5)] = 25

4x/5 + 1/2 [z + 1/4 (y + x/5)] = 45

3/4 (y + x/5) = 30

Simplifications:

4x/5 + 25 = 45

=> x = 5(45-25)/4 = 25

3/4 (y + x/5) = 30

=> (y + 25/5) = (30)4/3 = 40

=> y + 5 = 40

=> y = 40 - 5 = 35

1/2 [z + 1/4 (y + x/5)] = 25

=> [z + 1/4 (y + x/5)] = 2 (25) = 50

=> z + 1/4 (35 + 25/5) = 50

=> z + 1/4 (35 + 5) = 50

=> z + 1/4 (40) = 50

=> z + 10 = 50

=> z = 50 - 10 = 40

Conclusions:

Darius has 25 number of marbles, Ellen has 35 number of marbles, and Farhan has 40 number of marbles at the beginning.

There are 15 different ways.

h(t) = 8  ft

8 = -16t^2 + 50t+5

-16t^2 + 50t -3   = 0       Use quadrtatic formula    a = -16    b = 50     c = -3       

                                                       you will get two values for 't' at 8 ft    one is going up     one is coming back down

                                                               obviously the smaller value will be the first time the rock reaches 8ft ...on the way up

There are (5 - 1)^3 = 64 cubes that do not hav any black faces.

3 x 3 x 3 = 27 with no paint

6 choose 2 is actually 15, not 16, so the answer is 49

You should be able to work the area out easily from this picture

I got confused as I was doing this question but I think it is a trick question.

x^2-x cannot be 0 so  x cannot be 0 or 1  but I think it can be any other number

so I think that the domain is all real x where x is not equal to 0 or 1.

When the ball hits the ground , h = 0

so solve     0 = 1+15t -5t^2      for 't' using the Quadratic Formula    where a = -5    b = 15     c = 1       Throw out the negative 't' value

greatest height will occur at the 't' value = - b/2a        Use this value of 't' in the height equation to calculate the greatest height.

Find the least positive four-digit solution to the following system of congruences.
\(7x \equiv 21 \pmod{14} \\ 2x + 18 \equiv 16 \pmod{9}\)

\(\begin{array}{|rcll|} \hline 7x &\equiv& 21 \pmod{14} \\ 7x &\equiv& 21-14 \pmod{14} \\ 7x &\equiv& 7 \pmod{14} \\ 7x &=& 7 +14m \quad m\in \mathbb{Z} \\ 7x &=& 7 +14m \quad | \quad :7 \\ x &=& 1 +2m \quad | \quad \text{$x$ is an odd number!} \\ x &\equiv& 1 \pmod{2} \\ \hline \end{array}\)

...so \(7x \equiv 21 \pmod{14}\) or \(x \equiv 1 \pmod{2}\)

\(\begin{array}{|rcl|rcl|} \hline 2x + 18 &\equiv& 16 \pmod{9} \\ 2x &\equiv& 16-18 \pmod{9} \\ 2x &\equiv& -2 \pmod{9} \\ 2x &=& -2 +9n \quad n\in \mathbb{Z} \\ 2x &=& -2 +9n \quad | \quad x=1+2m \\ 2(1+2m) &=& -2 +9n \\ 2+4m &=& -2 +9n \\ 4m &=& -4 +9n \\ m &=& \frac{-4 +9n}{4} \\ m &=& \frac{-4 +8n+n}{4} \\ m &=& -1+2n+ \underbrace{ \frac{n}{4} }_{=r} \quad r \in \mathbb{Z} \\ m &=& -1+2n+ r & r &=& \frac{n}{4} \\ & & & 4r &=& n \\ & & & \mathbf{n} &=& \mathbf{4r} \\ m &=& -1+2(4r)+ r \\ \mathbf{m} &=& \mathbf{-1+9r} \\ \hline x &=& 1+2m \quad | \quad m=-1+9r \\ x &=& 1+2(-1+9r) \\ x &=& 1-2+18r \\ \mathbf{x} &=& \mathbf{-1+18r} \quad r \in \mathbb{Z} \\ \hline \end{array}\)

Find the least positive four-digit solution:

\(\begin{array}{|rcll|} \hline -1+18r &>& 999 \\ 18r &>& 1000 \\ r &>& \frac{1000}{18} \\ r &>& 55.\bar{5} \\ r &=& 56 \\ \hline x &=& -1+18*56 \\ \mathbf{ x} &=& \mathbf{1007} \\ \hline \end{array}\)

The least positive four-digit solution is 1007

This should help: