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Hi Wolfie,

I think you may have messed up some of the numbers. 

Woah, you stayed up quite late. 

=^._.^=

Hi Guest!

The x-intercept is 8, and the y-intercept is -10/3. 

I think Wolfie accidentally by 60 instead of 40 when finding them. 

x-intercept

x-intercept is when y = 0. 

5x - 12(0) = 40

5x = 40

x = 8

y-intercept

The y-intercept is when x = 0. 

5(0) - 12x = 40

-12x = 40

x = -40/12

x = -10/3

Distance

We can simply use a^2 + b^2 = c^2 since the x and y intercept form a right angle triangle. 

(-10/3)^2 + 8^2 = 676/9

sqrt(676/9) = 26/3. 

Answer

26/3

I hope this helped.  :)))))

=^._.^=

Concave up parabola

Focus(0,0)  A point (4,3)

Distance of the point to the focus is 5

Distance of (4,3) to the directrix is 5

So the directrix is  y=-2

So the vertex is (0,-1)

Equation is

\((x-0)^2=4a(y+1)\\ x^2=4(y+1)\\ so\\ y=\frac{x^2}{4}-1=\frac{x^2-4}{4}\)

\(4x+3y \le1000\\ 4x+3(\frac{x^2-4}{4})\le1000\\ 16x+3x^2-12\le4000\\ 16x+3x^2-12\le4000\\ 3x^2+16x-4012\le 0\\ find\;\;roots\\ x=\frac{-16\pm\sqrt{256+12*4012}}{6}\\ x=\frac{-16\pm 220}{6}\\ x=-39.\dot3\;\;or\;\;\;x=34\)

So this will be true for all integer values of x from -39 to  34  =  34- -39+1 = 74 integer x values.

BUT how many of these integer x values will have integer y values.

Only those where x^2 is divisible by 4.  If x^2 is divisible by 4 then x is divisible by 2. So x must be even.

So how many even numbers are there from -38 to 34 inclusive?   17+19+1 = 37

So 37 coordinate pairs will make this statement true.

Here is the graph.

A quicker way to look at it is that the difference between 15 and 20 is 5.

EF is 2/5 of the way from the short side.

2/5 of 5 = 2

so the length of EF will be 15+2 =17 units

Here is my take.

I have added a couple of points.

\(\frac{TE}{2x}=\frac{2.5}{5x}\\ TE=1\\ FE=1+15+1=17 units\)

Where did you get 30 degrees from?

What is the greatest common divisor(gcd) of \(2^{1001} - 1\) and \(2^7 - 1\)?

Formula: \(\boxed{ gcd(a^n -1,~a^m-1) = a^{gcd(m,n)}-1 }\)

\(\begin{array}{|rcll|} \hline a &=& 2 \\ n &=& 1001 \\ m &=& 7 \\ \hline gcd(a^n -1,~a^m-1) &=& a^{gcd(m,n)}-1 \\ gcd(2^{1001} - 1,~2^{7} - 1) &=& 2^{gcd(1001,7)}-1 \quad | \quad \mathbf{gcd(1001,7)= 7} \\ gcd(2^{1001} - 1,~2^{7} - 1) &=& 2^{7}-1 \\ \mathbf{gcd(2^{1001} - 1,~2^{7} - 1)} &=& \mathbf{127} \\ \hline \end{array}\)

Happy late new years!!

Happy new years!

Or for where I am!

That does not seem to be one of the answer choices. I forgot to include the answer choices with my post, here they are:

A) 8 B)10 C) 3√7 D)9 E)√65

OC = 32       CB = 40     OC + CB = OB       32 + 40 = 72 

And 

congrats on getting a amc 12 problem correct, the hardest problem on the test was correct,y answered by you. You are surely underrated

Yes pls explain lol

In the diagram, K, O, and M are centers of the three semi-circles. Also, OC = 32 and CB = 40. Line l is drawn to touch smaller semi-circles at points S and E so that KS and ME are both perpendicular to l. Determine the area of quadrilateral KSEM.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

EM = 20       SK = 52         KM = 72          SE = 8√65

EM and SK are the bases, KM is a side, and SE is the height of a trapezoid KSEM.

[KSEM] = 1/2(EM + SK) * SE = 1/2(20 + 52) * 8√65

[KSEM] ≈ 2321.93 square units

We have for the six faces the sum of the integers 1 through 7, inclusive, except for the one missing value m, and that sum is given to be 24. Therefore, 24 = 1 + 2 + 3 + 4 + 5 + 6 + 7 – m = 28 – m. So, m = 4, and the six faces are 1, 2, 3, 5, 6, 7, of which four values (2, 3, 5, 7) are prime, making the probability of rolling a prime to be 2/3. 

If you need a more in debt answer just ask

-wolfie

if I did get it wrong Hopefully someone can help you !

So guest

A radius of the biggest circle is OB   OB=OC+CB = 32+36 = 68 unitis

AO also equals 68 units.

AC= AO+OC = 68+32=100

So the radius of the middle sized circle is 50 units

KS=50

 -the area of trapezuim KSEM-

= average of the paralell sides * perpendicular height.

= 0.5( ME + KS) * ES

= 0.5( 18+ 50 ) * ES

= 34 * ES

SY=ME=18

KS = 50 = KY+YS = KY+18

so KY = 50-18=32 units.

Now KYM is a right angled triangle, t

The hypotenuse is KM = 50+18=68 units

one short side is KY = 32 units

KSEM = 34*ES = 34*60 = 2040 units squared. Im pretty sure

But I might have to double check my work but I think I got that right (I asked one of my class mates for help they said I was right)

If you need a more in debt answer just ask

-wolfie

That did not help and here's why:

If the sides are 4 and 16, then the perimeter is 40, and the area of a rectangle is only 64 square units.

Important: "The number of square units in its area is four times the number of units in its perimeter." 

a = 10        b = 40

Perimeter  P = 100

Area          A = 400  

 1250n+251250n+25 to 1250n+12251250n+1225. It turns out that only when n=0n=0 or 11 are the criteria satisfied, and in each case there are 2020 of the pairs. So 40 is the answer.

For more in debt answer just ask meh

-wolfie

(If wrong im sorry my brain hurts )

Wait I might've done that wrong Im not sure (sorry ive been answering so many people so my brain is gone right now)

Okay so lets start with

x intercept  witch is  (12, 0)

To  find the  y intercept, y  x  = 0 and that

12y  = 60

y = 5  ( I think)

  y intercept is  ( 0, 5)

The distance Would be-

  [ ( 12 - 0 )^2  + ( 5 - 0)^2 ]  =

 [ 12^2 + 5^2]  =

[ 144 + 25 ]  =

[ 169  ]  =  

13 units (If you need an more in debt answer feel free to ask)

-wolfie!

About finding general an+bn+cn,an+bn+cn, there is an elementary aspect to this, although not necessarily what you want. You already know that a+b+c=0,a+b+c=0, and the other answer gives enough to find a2+b2+c2a2+b2+c2 and a3+b3+c3.a3+b3+c3. Call those x1,x2,x3,x1,x2,x3, then solve in perpetuity with

xn+3=xn+1+xnxn+3=xn+1+xn

Two errors in the question,

bc+ca+ab=−1,bc+ca+ab=−1,

abc=1.abc=1.

So far, I get

x1=0,x1=0,

x2=2,x2=2,

x3=3,x3=3,

x4=2,x4=2,

x5=5,x5=5,

x6=5,x6=5,

x7=7,x7=7,

x8=10,x8=10,

x9=12,x9=12,

x10=17,x10=17,

Since the real root of x3−x−1x3−x−1 is slightly larger than 1,1, the numbers increase, are positive and so on. The two complex roots are smaller than 11 in modulus, so, for large n,n, we get xn≈Rnxn≈Rn where R≈1.3247R≈1.3247 is the real root. For example, R10≈16.643

For more in debt answer just ask (hopefully its debt enough through)

-wolfie

109.47122° or arccos(−​1⁄3)

Im not really good on this stuff but if you would want a more in debt answer just ask!

-wolfie

If you use 1+1+1−−−−−−−−√1+1+1, the answer is drilling straight through the cube.

(What a talented ant UwU)

However, we need the ant to go in a path on the surface of a cube, which poses new challenges.

We can quickly visualize the ant going up the cube while going around to the other side.

This, is a straight line along the 1×21×2 rectangle formed by the path.

Therefore, the shortest length is 12+22−−−−−−√=5–√12+22=5.

If you need a more in debt just ask!

-wolfie