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7  -  1 ==6 - this is the common difference.

Your sequence looks like this:

1 , 7 , 13 , 19 , 25 , 31 , 37 , 43 , 49 , 55 , 61 , 67 , 73 , 79 , 85 , 91 , 97 , 103 , 109 , 115 , 121 , 127 , 133 , 139 , 145 , 151 , 157 , 163 , 169 , 175 , 181 , 187 , 193 , 199 , 205 , 211 , 217 , 223 , 229 , 235 , 241 , 247 , 253 , 259 , 265 , 271 , >>Number of terms = 46>>Total Sum == 6256

6256 mod 6 ==4 - the remainder.

360/28 = 55/x

x = (28*55)/360

The perimeter of triangle ABC = 1/2(78) 

(a) The set of x that works is 1 <= x < 3/2.

(b) The set of x that works is 2 <= x < 8/3.

(c) The set of x that works is 3 <= x < 11/4.

(d) There are 58 possible values of x.

The perimeter of triangle ABC is 78/3 = 26.

y=3x^2+14x+9     This is a bowl shaped parabola because the +3 in front of x^2

                                  the minimum will occur at the vertex

                                         the 'x' value of the vertex is found by  - b/2a     or     -  14 / (2*3) 

                                               use this value of 'x' in the equation to calculate the 'y' value of the vertex......this is the minimum value of y

a piece of cloth must be cut to exactly cover a square table. the area of the table is 27ft squared. find the length of the side of the table to the nearest tenth of a foot.     

Wouldn't this just be the square root of 27?  Must we find it manually?   

I think I've forgot how to do that, so I'm going to use my calculator.   

So... sqrt(27)  =  5.196i52 which rounded to a tenth is 5.2 ft.   

_.

Good work Why.  I got the same answer.

Look under your original post.

Number in each small packet==S

Number in each large packet==L

L ==S  +  8

12S + 20[S +8] ==32S + 160 - total of all marbles.

1/4 * [32S + 160] =12S, solve for S

S ==10 - number of marbles in each small packet

L ==10 +  8==18 - number of marbles in each large packet.

(a)  18 marbles  x   20 large packets ==360 marbles.

(b) 360  + [10 * 12]==360  +  120 ==480 marbles altogether.

Look under your original post.

Let the number of pencils sold==P
The number of notebooks sold ==3/7P
Let the sale price of each pencil =S
The sale price of each notebook ==S + 4

3/7P * [S + 4]  - P*S ==120..........................(1)
3/7P * [S + 4] + P*S==750...........................(2), solve for P, S

Use elimination to get:
P==175 - pencils sold
S ==9/5 ==$1.80 - sales price of each pencil

(a) 175  x  $1.80 ==$315 - amount received from sale of pencils

(b) 175 + 3/7*175==175 + 75 ==250 - total pencils and notebooks sold.

(c) $1.80 + $4.00 ==$5.80 - cost of 1 notebook.

Sorry, that's incorrect

 First, determine the length c of the hypotenuse and then fill in the table of values.

Hello GamemasterX!

                        \(\ c\ =\sqrt{26}\)

     \(\ \delta\)                   \(\ A\)                               \(\ B\)

\( \ sin(\delta)\)      \(\dfrac{5}{\sqrt{26}}=\dfrac{5\sqrt{26}}{26}\)        \(\dfrac{1}{\sqrt{26}}=\dfrac{\sqrt{26}}{26}\)

\(\ cos(\delta)\)       \(\dfrac{1}{\sqrt{26}}=\dfrac{\sqrt{26}}{26}\)         \(\dfrac{5}{\sqrt{26}}=\dfrac{5\sqrt{26}}{26}\)

\(\ tan(\delta)\)                 \(\ 5\)                              \(\ \dfrac{1}{5}\)

  !

A  +  B  +  C==528.....................................(1)

A   -  3/8A  - 1/6A ==11/24A left

11/24A ==B + 3/8A...................................(2)

B  +  3/8A == C  +  1/6A............................(3), solve for A, B, C

A==384   and   B==32   and   C==112  - what the 3 boys started with.

a) Carl had 120 vs Bill who had 32

b)  See above

In the 12-hour period, the drawbridge is open: 8  *  15 ==120 minutes==2 hours.

Then probability is: 2 / 12 ==1 / 6

listfor(n, 1, 100, (-1)^(n+1)*(3*n)==

(3, -6, 9, -12, 15, -18, 21, -24, 27, -30, 33, -36, 39, -42, 45, -48, 51, -54, 57, -60, 63, -66, 69, -72, 75, -78, 81, -84, 87, -90, 93, -96, 99, -102, 105, -108, 111, -114, 117, -120, 123, -126, 129, -132, 135, -138, 141, -144, 147, -150, 153, -156, 159, -162, 165, -168, 171, -174, 177, -180, 183, -186, 189, -192, 195, -198, 201, -204, 207, -210, 213, -216, 219, -222, 225, -228, 231, -234, 237, -240, 243, -246, 249, -252, 255, -258, 261, -264, 267, -270, 273, -276, 279, -282, 285, -288, 291, -294, 297, -300)>>Total == - 150 sum of the sequence.

William had $108 more than Samuel. After William spent 3/4 of his money and Samuel spent half

 of his, both boys had a total of $102 left. How much money did each of them have at first?

S=Samuel,  W=William

Samuel ==S

W==108  +  S

1/4 * [108  +  S]  +  1/2S ==102, solve for S

S==$100 - what Samuel started with

W==$100 + $108 ==$208 - what William started with.

Can you explain 

This one is easy:

The first number whose digits sum is divisible by 3>>1,000,000,002

The last number whose digits sum is divisible by 3>>9,999,999,999

[9,999,999,999 - 1,000,000,002] / 3 + 1 ==3,000,000,000 - integers whose digits sum up to 3.

The roots are $1 + \sqrt{2} i$, $1 - \sqrt{2} i$, $-1 + \sqrt{2} i$, $-1 - \sqrt{2} i$.

The solution is (x,y,z) = (3,-1,2).

The area is 50*sqrt(3).