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This series is in arithmetic progression

with first no = N+1

Common difference = 1

(N)th term = 2N

Sum of N term = (N)/2 {2(N+1) + N-1}

= (N)(3N+1)/2

Average of this sum= (3N+1)/2

As per question

(3N+1)/2 = 20

3N+1 = 40

3N= 39

N= 13

Let triangle ABC have centroid G. D lies on AB such that AD/BD=2/3 and E lie on AC such that AE/CE =4/5. Lines DG and EG intersect BC at points P and Q, respectively. Find PQ/BC.

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I used an equilateral triangle and the answer is: PQ / BC = 7 / 12

Here's my work 

can you elaborate?

By mass points, PQ/BC = 5/13.

Points X, Y, and Z are on the sides QR, PR, and PQ respectively, of right triangle PQR  such that PZXY is a square. If PQ=12 and PR=8, then what is the side length of the square?

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Triangles PQR and ZQX are similar!

Square side => a

XZ / (PQ - PZ) = PR / PQ

a / (12 - a) = 8 / 12

a = 4.8

oh lord, this is hard to swallow 

oh well, lets start shall we -- pay attention to every step:

$\Huge\text{NOTE:}$  this website kind of screwed up the colouring i do not know why but check this link for a more accurate version: http://mathb.in/59340

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$ \color{lightskyblue}{\sqrt{60x}} \sqrt{12x}\sqrt{63x}\sqrt{42x} $

$  \color{lightskyblue}{\sqrt{4(15)x}}\sqrt{12x}\sqrt{63x}\sqrt{42x}  $

$ \color{lightskyblue}{ \sqrt{2^2(15)x}}\sqrt{12x}\sqrt{63x}\sqrt{42x}   $

$   \color{lightskyblue}{2\sqrt{(15)x}}\sqrt{12x}\sqrt{63x}\sqrt{42x}    $

now lets do the second one: $ 2\sqrt{(15)x}   \color{tomato}{\sqrt{12x}}\sqrt{63x}\sqrt{42x} $

$ 2\sqrt{(15)x}   \color{tomato}{ \sqrt{4(3)x}}\sqrt{63x}\sqrt{42x}  $

$ 2\sqrt{(15)x}  \color{tomato}{ \sqrt{2^2(3)x}}\sqrt{63x}\sqrt{42x}    $

$   2\sqrt{(15)x}   \color{tomato}{ 2 \sqrt{(3)x}}\sqrt{63x}\sqrt{42x}   $

now lets do the third one: $  2\sqrt{(15)x}2 \sqrt{(3)x}  \color{teal}{\sqrt{63x}} \sqrt{42x}     $

$   2\sqrt{(15)x}2 \sqrt{(3)x}  \color{teal}{\sqrt{9(7)x}} \sqrt{42x}    $

$   2\sqrt{(15)x}2 \sqrt{(3)x}  \color{teal}{\sqrt{3^3(7)x}} \sqrt{42x}    $

$   2\sqrt{(15)x}2 \sqrt{(3)x}  \color{teal}{(3\sqrt{(7)x}) } \sqrt{42x}    $

We are left with

$ 2\sqrt{(15)x}2 \sqrt{(3)x} (3\sqrt{(7)x})  \sqrt{42x}  $ 

firstly lets multiply the first two terms:

$ \color{darkorange}{2\sqrt{(15)x}2 \sqrt{(3)x}} (3\sqrt{(7)x})  \sqrt{42x}  $ 

$ \color{darkorange}{4\sqrt{(15)x} \sqrt{(3)x}} (3\sqrt{(7)x})  \sqrt{42x}  $

$ \color{darkorange}{4\sqrt{3x(15x)}} (3\sqrt{(7)x})  \sqrt{42x}  $

$ \color{darkorange}{4\sqrt{45x^2}} (3\sqrt{(7)x})  \sqrt{42x}  $

$ \color{darkorange}{4\sqrt{9(5)x^2}} (3\sqrt{(7)x}) \sqrt{42x}  $

$ \color{darkorange}{4\sqrt{3^2(5)x^2}} (3\sqrt{(7)x}) \sqrt{42x} \Leftrightarrow \color{darkorange}{4\sqrt{3^2 \cdot x^2(5)}} (3\sqrt{(7)x}) \sqrt{42x}  $

$\color{darkorange}{4\sqrt{(3x)^2(5)}} (3\sqrt{(7)x}) \sqrt{42x}  $

$\color{darkorange}{4[(3x)\sqrt{5}]} (3\sqrt{(7)x}) \sqrt{42x}  $

$  \color{darkorange}{12x\sqrt{5}} (3\sqrt{(7)x}) \sqrt{42x}  $

now lets do work the $12x\sqrt{5} (3\sqrt{(7)x})$ :

$\color{mediumseagreen}{12x\sqrt{5} (3\sqrt{(7)x})} \sqrt{42x}$  

$\color{mediumseagreen}{36x\sqrt{7x\cdot 5} } \sqrt{42x}$

$\color{mediumseagreen}{36x\sqrt{35x} } \sqrt{42x}$

finally, we multiply them all together:

$\color{navy}{36x\sqrt{35x}  \sqrt{42x}}$

$\color{navy}{36x\sqrt{42x(35x)}}$

$\color{navy}{36x\sqrt{1470x^2}}$

$\color{navy}{36x\sqrt{(7x)^2 \cdot 30 }}$

$\color{navy}{36x(7x\sqrt{30})}$

$\color{navy}{36x\cdot x(7\sqrt{30})}$

$\color{navy}{36x^2(7\sqrt{30})}$

$\color{navy}{252x^2(\sqrt{30})}$

or just 

$\color{navy}{252x^2\sqrt{30}}$

$\color{navy}{\left(6\sqrt{7}\right)x^2\sqrt{30}}$

$\color{navy}{6\sqrt{7\cdot \:30}x^2}$
 
$ \boxed{\color{navy}{ 6\sqrt{210}x^2}} $ 

can you do anything else to this? i dont think so.
 

that is just a little off.

x^2 + bx + c = 0

we know that one solution is -7, and the other one is 8

that means:

$\left. \begin{array}
 \ x=-7 \\ x=8 
\end{array} \right\}  \implies \left. \begin{array}
 \ x+7=0 \\  x-8=0 
\end{array} \right\} $    knowing that we can write the quadratic as a factorised form$ (x+7)(x-8)  $

new lets find the quadratic:

$  (x+7)(x-8) \overset{\left(m+n\right)\left(p-q\right) = mp-mq+np-nq }{=============\Rightarrow} xx+x\left(-8\right)+7x+7\left(-8\right) $

$xx-8x+7x-7\cdot \:8$

$ \boxed{ x^2-x-56 } $

‎‎‎‎‎‎‎

i feel like i have answered this before on a different website LOL. anyway.

we can set this up as an equation, by denoting the box as any variable -- i will let it be $n$

so we have $\left(7t-10\right)\left(5t+n\right) \overset{\left(t-m\right)\left(t+n\right)=t^2+nt-mt-mn }{==========\Rightarrow } 35t^2+7nt-50t-10n $

where 

$ y=\underbrace{35t^2}_\text{quadratic term}+ \underbrace{7nt-50t}_\text{linear term}  - \underbrace{10n}_\text{constant term} $

we can see that adding $n$ makes this a trinomial -- to turn it into a binomial we have to either work the constant term or the linear term by equalising them by 0.

1) constant term:  $ 10n=0 \implies n=0 $  ;  in our question it was stated that $n\neq 0 \ \  \wedge \ \  n\in \mathbb{Q}   $ so that does not statisfy our conditions

2) linear term:  $ 7nt-50t= 0 \implies  7nt=50t \implies n=\frac{50}{7} \overset{\text{as a mixed number}}{=======\Rightarrow} 7\frac{1}{7}  $ 

that does satisfy our conditions, so that is the value of $n$.

$ n=\boxed{7\frac{1}{7}}$

\(-560y^4\)

\(x^2+x-56=0\)

$ \log _2\left(24\right)-\log _2\left(3\right)=\log _5\left(x\right)\quad \overset{\Large{\text{if} \log_a(b)=c \: \therefore \: b=a^c }}{===============\Rightarrow }\quad \:5^{\log _2\left(24\right)-\log _2\left(3\right)}=x $

$5^{\log _2\left(\frac{24}{3}\right)}=x$

$5^{\log _2\left(8\right)}=x$

$5^{\log _2\left(2^3\right)}=x$

$5^{3}=x$

$\boxed{125=x}$

a1 = 1/4

r = 1/2

Sn = a1(1-r^n) / ( 1-r)

15/32 = 1/4(1- 1/2^n) /(1/2)

15/16 = 1-1/2 ^n

1/16 = 1/2 ^n

n= 4

Line l1 represents the graph of 3x + 4y = -14. Line l2 passes through the point (-5,7), and is perpendicular to line l1. If line l2 represents the graph of y=mx +b, then find m+b.

first of all lets find its perp line 

$3x + 4y = -14 \implies 4x-3y=$perp line where given the condition that that perp line should go through $(-5,7)$, respectively $x=-5$ and $y=7$:

$4\left(-5\right)-3\left(7\right)=x$
$-20-21=x$
$-41=x$

thus, the perp line which goes through $(-5,7)$ is $4x-3y=-41$ BUT the thing is we want that in the form of $y=mx+b$

knowing what we have is in the form of $Ax + By = C$ to find $m$, the slope, we have to do $\frac{A}{B}$ so $\frac{4}{3} $

the y intercep, or $b$ is $\frac{-C}{B}$ so $\frac{-(-41)}{3} \implies\frac{41}{3}  $

so the line in $y=mx+b$ form is $ y=\frac{4}{3}x+\frac{41}{3} $ 

Finally, lets find $m+b$:

$\frac{4}{3}+\frac{41}{3} =\boxed{\frac{45}{3}} \Rightarrow \boxed{15} $

here is an image of it as well:

Ummmm...1 +1 isnt possible i think..since i dont think  they put it back

Btw the answer is 30 but i dont know how to get to it :(

Switch the L side to log base 5

log5  24 / log 2     -    log5 3/ log2  

log5 (24/3) = log5 x

x = 24/3   = 8

At the x axis intercepts, the value of y = 0

First one:           0 = 1/3x + 7       x = -21

Second one :     0 = 2x+5            x = - 5/2                    sum = - 23 1/2

you couls mathematically solve them, or just use desmos

for $y = \frac{1}{3}x + 7$ the x value is $-21$ and for $ y = 2x+5$ the x value is $-2.5$

*What is the sum of the x-coordinates of the two lines' x-intercepts?*

$-21+-2.5=\boxed{-23.5}$

It would be nice if you said thank you.

You are writing solution to a different problem

Hey there, HT!

So...

Assume O to be the centre of triangle ABC, OT cross BC at M, link XM and YM. Let P be the middle point of BT and Q be the middle point of CT, so we have \(MT=3\sqrt{15}\). Since \( \angle A=\angle CBT=\angle BCT\) we have \( \cos A=\frac{11}{16}\). Notice that \(\angle XTY=180^{\circ}-A\), so \(\cos XYT=-\cos A,\) and this gives us \( 1143-2XY^2=\frac{-11}{8}XT\cdot YT.\) Since TM is perpendicular to BC, BXTM,  and CYTM cocycle (respectively), so \( \theta_1=\angle ABC=\angle MTX\) and \( \theta_2=\angle ACB=\angle YTM\). So \( \angle XPM=2\theta_1\), so \(\frac{\frac{XM}{2}}{XP}=\sin \theta_1\), which yields \(XM=2XP\sin \theta_1=BT(=CT)\sin \theta_1=TY.\) So same we have YM=XT. Apply Ptolemy theorem in BXTM we have \( 16TY=11TX+3\sqrt{15}BX\), and use the Pythagoras theorem we have \( BX^2+XT^2=16^2\). Same in YTMC and triangle CYT we have \( 16TX=11TY+3\sqrt{15}CY\) and \( CY^2+YT^2=16^2\). Solve this for XT and TY and submit it into the equation about \( \cos XYT\), we can obtain the result \( XY^2=\boxed{717}\).

Hope this helped! :)

( ﾟдﾟ)つ Bye

Hey there, Guest!

Greg: (x+1)(x+4) = x^2 + 5x + 5       \(c≠5 \)

Wilma: (x-1)(x-6)  =   x^2 -7x + 6.

Equation: \((x+3)(x+2) \)

Therefore the roots are -3 & -2.

Hope this helped! :)

( ﾟдﾟ)つ Bye

I have not done it but I think you just need to multiply top and bottom by  2-3i

and then expand and simplify.

Thanks Alan,

I was wondering about this one.