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We can set up variables to complete this problem. 

First, le'ts graph this problem. We have

Now, let's set up our first variable. 

Let's let the side of square be x. Since the base of the triangle is the same length as the side of the square, we have

base of triangle = x as well. 

Next, from the graph, which we nicely split into quarters, note that

height of triangle = \(\frac{3x}{4}\)

Now, since we have the base and height of the triangle, we can find it's area in terms of x. We have

\(\frac{3x}{4}\cdot x \cdot \frac{1}{2} = \frac{3x^2}{8}\)

The area of the square is just the side squared, so 

Area of square = \(x^2\)

Thus, we have what we need. 

\(ratio = \frac{3x^2}{8} \cdot \frac{1}{x^2} = \frac{3}{8}\)

So our answer is just 3/8. 

Thanks! :)

The rectangle with these coordinates is centered at the origin, with side length 2. Note that any equation y = mx will bisect the area of the square, meaning that essentially the line y=mx will actually go through the origin of the entire graph. 

As seen in the image below. 

Thus, the y intercept is essentially just 0.

Thanks! :)

We essentially have a system of equations to solve for x and y. The equation of

\( x^2 + y^2 = 1\\ y = x\)

Now, from the second equation, we know that x=y. Thus, subsituting out y for x, we have

\(x^2+x^2=1\\ 2x^2=1\\ x^2=1/2\\ x=\pm \frac{1}{\sqrt2} = \pm\frac{\sqrt2}{2}\)

Since x is equal to y, we have two points. We have

\((\frac{\sqrt2}{2},\frac{\sqrt2}{2}), (-\frac{\sqrt2}{2},-\frac{\sqrt2}{2})\)

Using the distance formula on these two points (or you could just graph it and count), we get

\(\sqrt{\left(-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\right)^2+\left(-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\right)^2}=2\)

So 2 is the answer. 

Thanks! :)

If angle XWZ = 60°, angle WXY = 180° - 60° = 120° because XY and WZ are parallels.

Angle ZXW must be less than 120°.

This problem is therefore invalid. 

Thanks! :)

Median length will be    ( 12   +     12+16) / 2 = 20 units 

As an aside, the height can be found by 

    area = median length * height 

      80 = 20 h 

        h = 4 units 

     (5 x 6 x 7 x 8) + 1  =  1680 + 1  =  1681    

_.

Every SEVENTH day is a day off for J ... L has every eigth day off...   every 6x8 = 48 days they are both 'off'.

365/48 = ~7 times a year       or possibly 8 times a year if they start off with the same day off at the early beginning of the year .

17b =   (1b + 7)

(1b + 7)^2  =  b^2 + 14b + 49

211b  = 2b^2 + 1b + 1

b^2 + 14b + 49  = 2b^2 + 1b + 1

b^2 - 13b - 48 = 0

(b - 16) ( b + 3)  = 0

b - 16  = 0

b = 16

\(4+ { \cfrac{1}{4 + \cfrac{1}{4 + \cfrac{1}{4}}}} \)

Work from the bottom-up

4 + 1/4 =  17/4

1/(17/4) = 4/17

4 + 4/17  = 72/17

1/(72/17)  = 17/72

4 + 17/72  =  305 / 72

2.57 repeating

[257 - 2 ] / [990 ] =  255 / 990 = 17 / 66

11012 / 100000   =   2753 / 25000

12 people can  count 1/4 box in  one hour

So 1 person  can count  1/ 4 * 1/12  =    1/48 box in  1 hr

So

30 people can count   30 * (1/48)  = 30/48 =  5/8  of a box in  1  hour

So

30  / (5/8) =   48 hrs

1/x - 1/(xy) - 1 / (xyz)  =  1/3

Let x  =1

1 - 1/3 =  1/(y) + 1/(yz)

2/3  = 1/y + 1/(yz)

2/3 = [ z + 1] / [ yz]

2yz = 3z + 3

2yz - 3z  = 3

z (2y - 3)  =  3

  If y = 2  then z  = 3

x,y,z = ( 1,2,3)

Check

1/1 - 1/2 - 1/6  =   1/3

1/977  =  .001........

2^4 * 3 =   48

4

1^2 =  1

2^2 =  4

P

  x =   3 + 2sin (2 *-pi6)  =  3 + 2 sin (-pi/3)  = 3 + 2  *(-sqrt 3 / 2) = 3 - sqrt 3

  y =    4 + 2cos (2 *-pi/6)  =  4 + 2 cos (-pi/3)  =   4 + 2(1/2) =  5

Q

x = 3 + 2sin (2*0)  = 3 + 2(0) =  3

y = 4 + 2cos (2*0) =  4 + 2 (1) = 6

R

x= 3 + 2sin (2 *pi/3) =  3 + 2 sin (2pi/3) = 3 + 2 (sqrt 3 / 2) =  3 + sqrt 3

y = 4 + 2cos(2 *pi/3) = 4 + 2 (-1/2) = 3

S

x = 3 + 2sin (2 *pi/2)  = 3 + 2 (0) =  3

y = 4 + 2cos (2 *pi/2)  = 4 + 2(-1) = 2

\(a \equiv 5 \pmod{7}\\ a+1\equiv 5+1 \pmod7\\ a+1 \equiv 6 \pmod7\)

6 is the answer

convert 2/7 to base 16

2/7_10 = 2/7_16 (nothing changes)

do base division

\(\frac{2}{7}_{16} \approx 0.4924924924924925\)

The answer is 49249