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Desmos is correct.

\(|x+3|=2 \\\text{should not be a slanted line because both} \\(x+3) \text{and}-(x+3) \text{is equal to 2} \)

To solve:

\((x+3)=2 \\ x=-1\)

\(-(x+3)=2 \\ -x-3=2 \\ -x=5 \\ x=-5\)

Two vertical lines, x = -1, and x = -5 is the graph of your equation. 

2)

4/12 = 1/3 Common ratio of the first GS

Sum = F / [1 - R], where F= First term, R = Common ratio

Sum =12/ [1 - 1/3]

Sum =12 / (2/3)

Sum = 12 x 3/2

Sum = 36/2

Sum = 18 - Sum of the first GS

18 x 4 = 72 - Sum of the second GS

72 = 12 / [1 - R]     divide both sides by 12

6 = 1 / [1 - R]         cross multiply

6 [1 - R] = 1           divide both sides by 6

[1 - R] = 1/6           subtract 1 from both sides

- R = 1/6 - 1

- R = - 5/6             Multiply both sides by -1

R = 5/6 - Common ratio of the second GS

12 x 5/6 = 60 / 6 =10 - this is the 2nd term of the second GS

10 - 4 = 6 value of n of the 2nd term of the second GS.

1) Sorry, Can't read the LaTex of your first question.

1)  a/(25 - a) + b/(65 - b) + c/(60 - c) = 7  This reduces to:

5/(a - 25) + 13/(b - 65) + 12/(c - 60) + 2 = 0   Add this to the 2nd equation in the question:

[5/(25 - a) + 13/(65 - b) + 12/(60 - c)] + [5/(a - 25) + 13/(b - 65) + 12/(c - 60)] + 2 = 2 - All terms cancel out leaving 2 as the answer.

SA of square pyramid =[Base Area] + 1/2 × Perimeter × [Slant Length ]      SA of square prism = 2 × Base Area + Base Perimeter × Length

Let the base area of both = S^2
The perimeter of both =4S
S^2 + [1/2*4S] * L = [2 * S^2] + 4S*S , solve for L=Slant Height
2 L S + S^2 = 6 S^2

Subtract S^2 from both sides:
2 L S = 5 S^2

Divide both sides by 2 S:
L = 5/2S - Slant Length[Height] of the pyramid.

I believe there's a simpler way for 1:

\(\frac{a}{25-a}+\frac{b}{65-b}+\frac{c}{60-c}=\frac{25-(25-a)}{25-a}+\frac{65-(65-b)}{65-b}+\frac{60-(60-c)}{60-c}=\\ (\frac{25}{25-a}-1)+(\frac{65}{65-b}-1)+(\frac{60}{60-c}-1)=\frac{25}{25-a}+\frac{65}{65-b}+\frac{60}{60-c}-3=\\ 5(\frac{5}{25-a}+\frac{13}{65-b}+\frac{12}{60-c})-3=7\\ \frac{5}{25-a}+\frac{13}{65-b}+\frac{12}{60-c}=x\Rightarrow5*x-3=7\Rightarrow5*x=10\Rightarrow x=2\)

There's nothing wrong with the question, there is sufficient information, look at it again.

Q1:

(6+r)² = 9² + r² 

6² + 2*6*r+r² =  9² + r²

6² + 2*6*r =  9² 

2*6*r =  9² - 6²

r = ( 9² - 6² ) / ( 2*6 )

Q2: 

by using pythagoras rule

if    RT² = RS² + ST²    is true , ST is tangent to the circle.

         (5+8)²   =    5² + 12²

           (13)²   =    5² + 12²

            169    =    25 + 144

            169    =    169

yas 

Both questions begin by making use of the Pythagoras theorem.

For the first one, start by saying that 9^2 + r^2 = (6 + r)^2, tidy up and solve for r.

The second one is asking you if the triangle is rightangled (Pythagoras again), if it is then ST will be a tangent.

Q1:

X= 3/tan(30)

Q2 :

a= 8*sin(60)

b= 8*cos(60)

small Triangle area (At) = (a+b)/2

Total Area = At*2*6

Q3: 

X= 4.2/tan(45)

Area = 2*8*X+8²

Q4: 

X= 4.2/tan(60)

Area = 2*8.4*X+8²

I'm pretty sure it is unanswerable with that limited info, but I am rusty on this topic.

Remember your 30-60-90 and 45-45-90 shorctcuts!

First One: 3√3

If you do not understand one of the steps, explain what your problem is the best you can and then I, or someone else, will help you more :)

\(y=x^2-6x+5\\ x^2-6x=y-5\\ x^2-6x+9=y-5+9\\ (x-3)^2=(y+4)\\ vertex(3,-4)\)

There is a “Chicken McNugget’s Theorem,” but there should not be one.

The real reason is the “cute” name “Chicken McNugget’s Theorem.”  If it’s “cute” then more will remember it and use it. Cuteness in general, and anthropomorphic cuteness in particular, is the byword for educating children and adults too.  We might not understand it unless a blòódy dancing McNugget explains it to us.

Solving two variables in a single equation is usually a trial and error process. However, in this case, you can easily narrow the process to a few guesses using logic.

The (i) and (j) values are integers; there are no fractional or decimal components in them.

Because the sum of integers is very low, it reasonable to assume the (i,,j) exponents are low—less than 10.

To solve, start by assuming a central value (between 1 and 10) for (j), and then use logarithms to isolate for (i).  

Because (2) is one of the factors, its resolution will be its multiples 2,4,8,16,32,64,128,256, ... ect. Seeing one of these indicates you’ve found the solution set.

\((( 2^i-1)*\dfrac{(3^j - 1)}{(2)} = 600 \leftarrow \tiny \text{ The (2) and (1) are common to all equations.}\\ ( 2^i) = \dfrac{1200}{(3^j - 1)} +1\leftarrow \tiny \text{ Leave it in this form to streamline the solution search.}\\ \text {Example: assume (j) = 6 and solve for (i)}\\ ( 2^i) = \dfrac{1200}{(3^6 - 1)} +1\\ ( 2^i) = \dfrac{150}{91} +1\\ Log_2( 2^i) = Log_2 \left( \dfrac {150}{91} +1\right) \small \text{ Note the fraction; This is not the solution.}\\ \text {Example: assume (j) = 5 and solve for (i)}\\ ( 2^i) = \dfrac{1200}{(5^5 - 1)} +1\\ ( 2^i) = \dfrac{600}{121} +1\\ Log_2( 2^i) = Log_2 \left( \dfrac {600}{121} +1\right) \small \text{ Again, note the fraction; This is not the solution.}\\ \text {Example: assume (j) = 4 and solve for (i)}\\ ( 2^i) = \dfrac{1200}{(5^4 - 1)} +1\\ ( 2^i) = 15 +1\\ Log_2( 2^i) = Log_2 \left( 16\right) \small \text{ Note the integer 16, corresponding to i=4; This is the solution.}\\ \)

The solution is j=4 and i=4.  Subtract (1) from each and use these as the exponents for the primes 2, and 3 to find the number (N) for which 600 is the sum of its divisors.

There are similar techniques for (N) that have three or more unique prime factors.

GA

Edit: Corrected to indicate base (2) log.

Here is the full solution.

Just as guest said.

1. 

\(\frac{a}{25-a}+\frac{b}{65-b}+\frac{c}{60-c}=\frac{25-25+a}{25-a}+\frac{65-65+b}{65-b}+\frac{60-60+c}{60-c}\)

\((\frac{25}{25-a}-1)+(\frac{65}{65-1}-1)+(\frac{60}{60-c}-1)=\frac{25}{25-a}+\frac{65}{65-b}+\frac{60}{60-c}-3\)

After factoring out the five, we have:

\(5(\frac{5}{25-a}+\frac{13}{65-b}+\frac{12}{60-c})-3=7\)

We can see that the value of what we are trying to find and its relantionship to the original expression.

Since the original expression is 7, and the expression we are trying to find is 3 less than 5 times 7,

We have:

\((7+3)\div5=2\)

Just the way guest did it!

I hope this helped,

Gavin.

LOL, Let it be. XD

That is NOT what it says in English!!. You better remove it.

I do. Did you understand my pfp? XD

quilly: Do you speak Arabic language?

1.   16(x - 3)  =  (y - 1)^2 

The vertex of this parabola is at  ( 3, 1)

This parabola opens to the right

In the form    4p(x - 3)  = (y - k)^2......the focus is  ( 3 + p, 1)   

We can solve for p as  4p  =16  ⇒   p  = 4

So....the focus  is  ( 3 + 4, 1)   =  (7,1)

The dirctrix lies 4 units to the left of the vertex and is given by    x  = ( 3 - p )  =   (3  - 4 )   =    -1

Here's a graph :  https://www.desmos.com/calculator/iwubncnnan

2.  y  = -1/6 x^2 + 7x  - 80     we need to complete the square on  x

y  = (-1/6)  [ x^2- 42x + 480 ]     

Take 1/2 of 42  = 21....square it  = 441....add and subtract it within the brackets

y   =    (-1/6)  [x^2 - 42x + 441 + 480  - 441 ]     factor the frist three terms....simplify the rest

y  = (-1/6) [  (x - 21)^2  + 39 ]

y = (-1/6)(x - 21)^2  - 39/6

y  = (-1/6)(x - 21)^2  - 13/2       add 13/2 to both sides

(y + 13/2) =  (-1/6)(x - 21)^2        multiply through  by -6

-6(y + 13/2)  = (x - 21)^2

Because of the presence of a "-"   in front of the "6"......this parabola turns downward

The vertex is  ( 21, - 13/2)

The focus is  given by ( 21 , -13/2  + p )

And we can find  p as    .....  4p  = -6   ⇒  p  = -6/4  = -3/2

So....the  focus  is  (21, -13/2 - 3/2)   = ( 21 , -16/2)  = (21, -8)

The directerix is given as    y  = (-13/2 - p)   = ( -13/2 - (-3/2) )   = ( -13/2 + 3/2)  = -10/2  = -5

Here's a graph :  https://www.desmos.com/calculator/g9zfw92yos

thank you so much CPhill you always help me so much!

1.  y  = -1/4x^2 + 4x  - 19

The x coordinate of the vertex  is given by   -4 / (2 * -1/4)  =  -4 / (-1/2)  =   -4 * -2    =  8

And the y coordinate is given  by :

-1/4 (8)^2  + 4(8)  - 19  =

-16  + 32   - 19

-3

So....the vertex  is  (8, -3)

Here's a graph : https://www.desmos.com/calculator/z76eahtb61

2. The center of the circle  is  ( 1, 4)   and the radius  is 8

So we have

(x - 1) ^2   + ( y - 4)^2   = 8^2        expand and simplify

x^2 - 2x + 1  + y^2  - 8y + 16  = 64

x^2 + y^2  - 2x - 8y  + 17  = 64       subtract 64 from both sides

x^2  + y^2  -2x - 8y - 47   =  0 

3.  We have    y   =  1/8 x^2 + 4x + 20

We want to complete the square  on x

y = (1/8)  (x^2  + 32x + 160)

Take  1/2 of 32  = 16....square it  = 256....add and subtract it within the parentheses

y  = (1/8)  (x^2 + 32x  + 256   + 160   - 256)       factor the first three terms, simplify the last two

y = (1/8) [ (x + 16)^2  - 96 ]       apply the  1/8   over both terms in the parentheses

y  = (1/8) (x + 16)^2 - 12

(y + 12)  = (1/8)(x + 16)^2     multiply both sides by 8

8( y + 12)  = (x + 16)^2

The vertex  is at  (-16, -12)

In the form

4p ( y - k)  = (x - h)^2

4p  = 8

So....p  = 2

And the focus is given by

(-16 ,  -12 + p)   =    (-16, -12 + 2)   =   (-16, -10)

Here's the graph  :  https://www.desmos.com/calculator/cgsvcjy0ny