TaliaArticula

Yikes, sorry, HT.

Hey there, Guest!

\(x^2+10x+25=8+17\)

Step 1: Simplify both sides of the equation.

\(x^2+10x+25=25\)

Step 2: Subtract 25 from both sides.

\(x^2+10x+25−25=25−25\)

\(x^2+10x=0\)

Step 3: Factor the left side of the equation.

\(x(x+10)=0\)

Step 4: Set factors equal to 0.

\(x=0\) or \(x+10=0\)

x=0 or x=−10

Therefore your answer is -10.

Hope this helped! :)

( ﾟдﾟ)つ Bye

Hey there, Guest!

Let's find the critical points of the inequality:

\(x^3+4x=5x\)

\(x^3+4x−5x=5x−5x\) (Subtract 5x from both sides)

\(x^3−x=0 \)

\(x(x+1)(x−1)=0\) (Factor left side of the equation)

\(x=0\) or \(x=-1\) or \(x=1\)

Check intervals in between critical points. (Test values in the intervals to see if they work.)

\(x<−1\) (Doesn't work in original inequality)

\(−1  (Works in original inequality)

\(0  (Doesn't work in original inequality)

\(x>1\) (Works in original inequality)

Therefore the answers to your equation are: 

−1  or  x>1

Hope this helped! :)

( ﾟдﾟ)つ Bye

Hey there, Guest!

1x + 2x + 3x + 4x + 5x + 6x = 1

Therefore 21x equals 1.

x = \(\frac{1}{21}\)

Hope this helped! :)

( ﾟдﾟ)つ Bye

Hey there, CCeed!

Ahaha, yerpp, you're using this website correctly; thank you for creating an account so that I have someone to address as besides "Guest". 😂

1. Let's see what all the factors of 16 are: 1, 2, 4, 8, and 16

1 & 16 are a pair, 2 & 8 are a pair, and 4 & 4 are a pair.

1+16=17

2+8=10

4+4=8

x=8 and y=2 or the other way around.

Therefore the answer is 6 or -6.

As for the other two questions, I don't know how to solve them correctly...

Hope this helped! :)

( ﾟдﾟ)つ Bye

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

Hey there, Guest!

So...

An n number is a 1 followed by 2^(n-1) combinations of n-1 0's and ones, symmetrically distributed, hence (n-1)2^(n-2) zeros.

In total,

1+2*2+3*4+4*8+5*16+6*32+7*64+8=777.

so that for numbers between 0 and 2^n, the number of zeroes is

(n−2)2^(n−1)+n+1

Hope this helped! :)

( ﾟдﾟ)つ Bye

Hey there, Guest!

The answer is 66 ways.

And here is the explanation:

Hope this helped! :)

( ﾟдﾟ)つ Bye

Hey there, ICL!

So, sometimes when you can't solve an equation, you can plug it into a Desmos Graphing Calc and then you'll be able to see the vertex.

EP is right about this one, x=2 and y= -7; here's the link to the graph: https://www.desmos.com/calculator/9nnq18slcn.

Hope this helped! :)

( ﾟдﾟ)つ Bye