All Answers 358089 Answers

Here's my answer :   https://web2.0calc.com/questions/coordinates_42876

are u for real rn

like mate ngl you capping or...

its 32=8+8+8+8

\(\triangle AXB \) is Isosceles so \(\angle XAB = \angle XBA\)
\(\angle DAB=\angle CBA=90^\circ\)
so \(\angle DAX=\angle DAB-\angle XAB=\angle CBX\) thus 
\(\triangle DAX\cong\triangle CBX\) (SAS SIMILARITY, \(DA=CB\))
so \(DX=CX=\boxed{5}\) (by CPCT)
 

Note that  if f(x)  = x^2   we have

(x^2 - 2x - 5) * x^2 =  x^4 - 2x^3 - 5x^2

So

k =  -5

f(x) =   2 / ( 2 - 3x)

Range :   We have  [ lower power polynomial / higher power polynomial]....when this happens  we have a horizontal asymptote   at  y =  0

So, the range is  [ -inf, 0) U (0, inf)

Domain :  Whatever x value makes the denominator =  0  is NOT  in the  domain

So

2- 3x  =  0

2= 3x

x = 2/3

Domain =  (-inf , 2/3) U ( 2/3 , inf)

We have the form

y =  A cos  (B (x - D)) + C

A = the amplitude =   [ 3 - - 11] / 2 =  14 / 2 =  7

C= the midline =  [ 3 - 11 ] / 2 =   -8 / 2 =   -4

The period  =  [ ((5/2)pi + 2 pi)) / 2    - pi/4 ] =   [ (10/4)pi + (8/4)pi ) / 2   - pi/4]  = 

[ (18/4)pi / 2 - pi/ 4] =  [ (18/8)pi - pi/4 ]  = [ (9/4)pi - pi/4 ] = (8/4)pi  = 2pi

B =  2pi / period =  2pi / 2pi  = 1

D = the phase shift = pi/4 

The equation is

y = 7cos ( x  - pi/4) - 4

See here  :

We have the form

f(x)  = A sin (Bx) + C

A = the amplitude.....the (-)  reflects the graph over the x axis

C =  the vertical shift upward =  4   (the midline)

B =  2pi / period =   2pi /  2  =  pi

The graph is

f(x) = -10sin (pi x) +  4

Let line p= \(y=0\) and line q= \(x=y\) this satisfies that the lines have an angle of \(45^\circ\) between them and they meet at \(X=(0,0)\)

If point \(A\) lies on line p then its refelction across the line will be itself. So \(A=B\)

Now for the relection of point \(A\) across the line q.

WLOG let point \(A\)'s distance from \(X\) be \(1\)
Then \(A\)'s perpendicular distance from line q will be \(\frac{1}{\sqrt{2}}\). Reflecting means we find point \(C\) to have the same perpendiclar distance to q from the other side.
We notice that \(C\) lies on the lies on the y axis and \(A=B \) lies ont he x-axis making the \(\boxed{\angle BXC=90^\circ}\)

ggwp

Here :  https://web2.0calc.com/questions/angles_103

Amplitude =   [ greatest y value on graph - least y value on graph ]  / 2  =

[ 4 -  - 4 ] /   2   = 

[8] /  2 =  

4

AC is the diagonal  = 2

x is the side of the square =  diagonal / sqrt 2 =   2 / sqrt 2 =  sqrt 2

Sry but,

I just saw the answer in the link you have sent and I got a different answer. So here is what I did:

23 is the smallest prime \(>20\) and all different primes till 83 satisfy the condition. This is 14 primes

similarly 29 matches with 9 primes(till 67), 31 with 6 (till 59), 37 with 4(till 53), and lastly 41 with 2(till 47).

This is a total of \(\boxed{35}\) different products as all the numbers have a different pair prime factors.

This is the graph of the sec  whose period is  twice as long with an amplitude of 3  shifted up one unit

See here : (the sec is in red....our graph is in blue )

Here :  https://web2.0calc.com/questions/help-plz_85848

There are 4 vowels and the rest 12 are consonants. 
There are 4 cases:

CASE 1: A
If one picks A as the first letter then there are 15 more choises to pick.

Case 1 has 11 choises

CASE 2: E
the second letter must be anything starting from F meaning 11

CASE 3: I
there are 7 choises

CASE 4: O
there is only one choice which is P

Therefore
\(15+11+7+1=\boxed{34}\)
(NOTE: this is not counting AA, EE, II, AND OO. if so just add 4 to 34= 38)
ggwp

\(\triangle XYD\) has base \(XY=12\) and Hmax of:

\(\frac{1}{2}bh=10\\ h=\frac{5}{3}\)
Thus the desired area is \(\frac{5}{3}\) from base \(XY\) (A straight line parallel to \(XY\))
Let the line touch \(YZ\) at \(A\) and \(ZX\) at \(B\) 
\(P=\frac{\text{desired area}}{\text{total area}}\)

So: Total area=\(\frac{1}{2}bh=36\) and desired area is:

\(\frac{1}{2}h(XY+AB)=\frac{155}{9}\)  (\(AB=\frac{26}{3} \) due to similar triangles \(\triangle ZAB \sim \triangle ZYX\))

Thus: \(\boxed{P=\frac{155}{324}}\)

ggwp :)

Lets start by assembling the beads in a staight line:

There are \(\frac{7!}{5!}=42\) number of ways to arrange them in a straight line (\(7!\) for the number os beads and divide by \(5!\) for the number of same beads).
Now to divide due to rotation and reflection:

To do this we divide by 7 for the number of rotations and by 2 for the mirror image:

therefore:

\(42/14=\boxed{3}\)

ggwp

We can do this by casework:

Case 1: CORNERS

Each corner has \(5\) circles we can choose that are not next to it (The 5 on the border). There are 4 corners so, \(4\times5=20\)

Case 2: EDGES

Each edge has 3 circles we can choose that are not next to it (The 3 opposite to it). There are 4 edges so, \(4\times3=12\)

so: \(20+12=\boxed{32}\)

Note: there woudve been a third case with the center but the center is next to all the cricles in a 3x3 grid

ggwp :)

We know \(AD=BA=3+2=5\) 

\(AQ=AP= \sqrt{AD^2+DQ^2}=\sqrt{34}\)

\(\triangle AQD \cong \triangle ABP\) by SSS congruence
Thus:
\(\angle QAD=\angle PAB\) (by congruence above)
And \(\angle PAQ=90-2\angle PAB\) 
 Let \(\angle PAB =x\)and \(\angle PAQ=90-2x\)    (for simplicity)
 

\(\sin(90-2x)=\cos(2x)=1-2\sin^2x\)

We know \(\sin x=\frac {PB}{AP}=\frac{3}{\sqrt{34}}\)
so finally:

\(\sin {\angle PAQ}=\boxed{\frac{8}{17}}\)
ggwp

So, here is my best shot:

Let \(f(f(x)=y\)  and  \(y\in {1,2,3,4}\)

Therefore:

\(f(f(x))=f(f(f(x)))\) becomes \(y=f(y)\)

This means we just need to find the probability of \(f(x)=x\) and the value of \(f(f(x))\) can be anything, it doesnt matter.
Thus, the probability of

 \(f(x)=x\) where \(x\in {1,2,3,4}\)
\(= \boxed{\frac{1}{4}}\)

hope its right ngl

We move everything to one side, \({n}^{2}+40n-144<0\).

We don't see a good way to factor, so we apply the quadratic formula, \(n_{1, 2} = -20 \pm 4\sqrt{34}\). This is also \(n_{1, 2} = -20 \pm \sqrt{544}\). The solutions shouls be in between these two roots because a > 0 in this quadratic. Square root of 544 is a little more than 23, so we know in the range, there is -20, 23 integers greater than -20, 23 integers less than -20. So there is 23*2 + 1 = 47 total integers in this range.

x(x + 6) > 16 - 7x + 20

x^2 + 6x + 7x - 36 > 0

x^2 + 13x - 36 > 0

x^2 + 13x  > 36     complete the square on  x

x^2 + 13x + 169/4 > 36 + 169/4

( x + 13/2)^2  > 313/4

x + 13/2 >   sqrt (313)  / 2               x + 13/2 < -sqrt(313)/2

x >  [ sqrt (313) - 13 ]  / 2                x <  [ -sqrt (313) - 13 ] / 2

Solution

(-inf , [ -sqrt (313)  -13 ] / 2 )   U  ( [sqrt (313 - 13 ] / 2 ,   inf )