All Answers 352599 Answers

Rewrite the expression \( 6j^2 - 4j + 12\) in the form \( c(j + p)^2 + q\), where c, p, and q are constants. What is \(\frac{q}{p}\)

\(​​​​6j^2-4j+12\\ =6(j^2-\frac{2}{3}j)+12\\ =6(j^2-\frac{2}{3}j+\frac{1}{9})+12-\frac{6}{9}\\ =6(j-\frac{1}{3})^2+11\frac{1}{3}\\ =6(j-\frac{1}{3})^2+11\frac{1}{3}\\ =6(j-\frac{1}{3})^2+\frac{34}{3}\\~\\ p=\frac{-1}{3}\qquad q=\frac{34}{3}\\ \frac{q}{p}=\frac{34}{3}\div\frac{-1}{3}\\ \frac{q}{p}=\frac{34}{3}\times\frac{3}{-1}=-34\\ \)

fyi, based on my patterns above, it equals 957 too

33+42+63+84+105+126+147+168+189 = 957

for X, 2/3 is bigger than 1/2, so that 1/2 should go up to be subtracted from 2/3,

but for Y, 1/2 is bigger than 1/4, so that 1/4 should go down to be subtracted from 1/2.

with the X's on top and the Y's on the bottom you get.

X^2/3-1/2

_______

Y^1/2-1/4

hopefully you recall how to find common denomiators for the top.

2/3 - 1/2

4/6 - 3/6 = 1/6

logic will tell you if you have half and subtract a quarter, a quarter remains,

if not then do the common demoniators again

X^1/6

_______

Y^1/4

How many integers between 200 and 300 have the sum of their digits equal to 15?

First decide how many combinations of 2 digits make 13

So the only numbers are 

249

294

258

285

267

276

So that makes  6 integers

Summed up on Wolfram/Alpha and it gives:

57 + sum_(b=0)^9( sum_(a=0)^9(a + b)) =957

https://www.wolframalpha.com/input/?i=57+%2B+%E2%88%91%5Ba%2Bb%5D,+a%3D0+to+9,+b%3D0+to+9

Here is an extremrly basic mockup of a number table from 0-99 for you to reference as I make the adding process easier.

I notice a few patterns here. The tens digit remains the same. Let's attempt to add the numbers 0-9.

\(\underbrace{0+\underbrace{1+\underbrace{2+\underbrace{3+\underbrace{4+5}+6}+7}+8}+9}\\ \)

The numbers enclosed by the underbraces sum to nine (9+0=9, 1+8=9, 2+7=9, 3+6=9. 4+5=9). There are 5 lots of this occurring with the numbers one to nine, so the sum of the tens digit in the first columns equals \(9*5\text{ or }45\). This phenomenon occurs ten times or once per column, so let's multiply this by 10. \(45*10=450\). 

Now, let's do the ones column. The ones column is different because each column increments the amount by 1. The sum of the ones column can be represented by the following sequence. \(0*10+1*10+2*10+...+8*10+9*10\). We can use algebra to group the common factor, so it simplifies to \(10(1+2+3+...+8+9)\). Of course, we already know that the sum of the numbers 1-9 sums to 45 by our first calculation. 10 times that amount yields \(450\). 

Of course, we have only dealt with the sum of the digits from 0-99. However, if you think about it, the sum of 100-109 should be the same as adding the ones digit and the hundreds digits. Of course, this is a special case since the tens digits are all 0. We already know that \(1+2+3+...+8+9=45\). We know that the hundreds digit of the numbers 100-109 sums to \(1*10=10\). However, we have forgotten 110, which has a value of 2. \(45+10+2=57\) for the sum of the digits from 100-110. Now, let's add everything together.

\(450+450+57=957\) or the sum of the digits from 0-110

1) Find the area of a regular 12-gon inscribed in a unit circle.

We have 12 identical triangles  identical isosceles triangles with equal sides of 1 and whose apex angle  between these sides =  360/12  = 30”  

The  area will  be given  by    (1/2) (1)^2*sin (360/12°)  =

(1/2 sin (30°)  =  1/2  *  1/2  =   (1/4 ) units^2

2) A regular hexagon has side length 6. If the perimeter and area of the hexagon are p and A, respectively, what is the value of (p^4)/(a^2)?

The perimeter, p, is 36  ⇒   p^4  =   36^4

The area, a,  is   (1/2)6^2sin (60)  =  18*√3/2  =  9√2  ⇒ a^2  = (√182) ^2   = 162

So p^4 / a^2  =    36^4  / 162   =   10368 

3)Isosceles triangle OPQ has legs OP = OQ, base PQ = 2, and and angle POQ = 45 degrees. Find the distance from O to PQ.

The distance from O to PQ  is the altitude....call the point where the altitude intersects the base, R

And this altitude bisects POQ....so angle POR  =  22.5°

And the altitude also bisects the base.... so PR  =  1

Using the tangent function....we have that

tan (22.5)  =  1 / altitude        rearrange as

altitude  =   1  / tan (22.5)  ≈ 2.4142  =    1 + √2   

4) A, B, C, D and E are points on a circle of radius 2 in counterclockwise order. We know AB = BC = DE = 2 and CD = EA Find [ABCDE].
Enter your answer in the form x + y√z in simplest radical form.

This is a regular pentagon inscribed in a circle....I assume you want the area of ABCDE??

If so...... the area is   5 (1/2)(2)^2sin (72)  = 10sin (72)  = 10 √  [  5/8  + √5/8 ]  units^2

If you want the perimeter...we  have that  the half side lengh  = 

2sin(36)

And we have 10 half side lengths comprising the perimeter....so....the perimeter  =

10 * 2 *  sin(36)   = 20√ [ 5/8  - √5/8 ] units

The method used by Guest #2 works ONLY for numbers all ending in 9s. Therefore, the following is the more accurate method:

0, 1, 2, 3..............96, 97, 98, 99. Will add first and last=0 +99 =18

Will add second to second last =1 + 98 =18........and so on.

Since we have: 100/2 =50 pairs, each equaling 18, we therefore have:

50 x 18 =900

The only numbers left 100 to 110 itself, and since they have:1+2+3+4+5+6+7+8+9+10+2=57

900 + 57 =957 - Sum of all digits from 1 to 110.

249, 258, 267, 276, 285, 294 = 6 numbers.

👍  :D

i only have two chances if i get it wrong 2 times ill have to start all over again with  new problems... tough work

which one should i try?

0, 1, 2, 3..............106, 107, 108, 109, Will add first and last=0 +109 =10

Will add second to second last =1 + 108 =10........and so on.

Since we have: 110/2 =55 pairs, each equaling 10, we therefore have:

55 x 10 =550

The only number left is the 110 itself, and since it has a value of 2, we therefore have:

550 + 2 = 552. And that is what I think !!.

i think you would just count up the 1's then the 2's and so on, then multiply and add.

so there is 1, eleven 1's in 10-19, eight 1's in 20-99, thirteen 1's in 100-110,

1+11+8+13 = 33

so there are 2, 12, and 102,  then eleven 2's in 20-29, seven 2's in 30-101.

(3*2) + (11*2) + (7*2) = 42

i see some patterns, but test it out by doing the 3's and 4's?

(4*3) + (11*3) + (6*3) =63  based on patterns

would the 4's = 84? test and see

can't you just replace x with 180x/Pi?

Well . . . . What do you know? You told us where the @#$* S and F are. How sweet!

You could have just appended it to your previous post.

4x^2 +  2x  - 1        complete the square on x

4 (x ^2  +  (1/2)x   -   1/4 )

4( x^2  + (1/2)x + 1/16  -  1/4  -  1/16 )

4 (   (x + 1/4)^2   -  5/16 )

4(x + 1/4)^2  -  5/4

a  = 4      b  =  1/4       c  =  -5/4

a  +  b   +   c   =      4 +  1/4  - 5/4     =   3

x ( x - 3)  =  1     write as

x^2  - 3x   -  1   =   0             complete the square on x

x^2  -  3x  + 9/4    -  1  -  9/4   =   0

(x - 3/2)^2  -  13/4  = 0

(x - 3/2)^2   =   13/4       take both roots

x -  3/2   =  ±√(13/4)

x -  3/2  =  ±√13 / 2

x  =   [ 3 ±√13 ] / 2

abc  =  3 * 13 *  2  =  78

Oh...Ok    sorry....    If you multiply all of my x.y 's together (rounded)  they come out to -7       

Nice, EP!!!

18 + 5x^2  =  20x     rewrite as

5x^2 -  20x +  18  =  0

Complete the square on x

5 ( x^2   -  4x  +  18/5)  = 0

5 ( x^2  -  4x   +  4   +  18/5  -  4)  = 0

5 (  (x - 2)^2   +  18/5  - 20/5 )   =   0

5( (x - 2)^2  - 2/5)  = 0

(x - 2)^2  -  2/5  =   0

(x - 2) ^2   =  2/5           take both roots

x -2  = ±√(2/5)

x  =    2 ±√(2/5)

This evaluates to     x  ≈  1.3675    x  ≈ 2.6325

The desired rounding produces     1   and 3     and their product is   3

3x^2 + x - 4   factor out the 3

3( x^2 + (1/3)x  -  4/3 )

Take 1/2 of 1/3  = 1/6  .....square this  = 1/36....add and subtract it within the parentheses

3(x^2  + (1/3)x  + 1/36   -  4/3  -1/36 ) 

Factor the first three terms in the parentheses and simplify

3  (   [ x + 1/6 ]^2  -  48/36  -  1/36 )

3 ( [ x + 1/6 ] ^2  - 49/36 )

3(x + 16)^2 -  49/12 

So.... k  =  -49/12

No problem ACG....I wouldn't expect anyone to scroll through the thousands of questions on here to see if theirs has been asked before.....I just happened to remember this one....to give full credit, I believe that heureka  or alan may have solved it originally....

Thanks, alan...I like your method better.....!!!