All Answers 354232 Answers

Good job, Swag  !!!

Every quadrilateral has 360 degrees

1 and 2 both have equal angles, so we can add 48 and 90 to get 138

Subtract 138 from 360 to get 222

Now we have to divide 222 in half, since we need to find the angle of one of the numbered angles

222/2 is 111

We can check this by adding all of the angles together and making sure that they equal 360

111+111+48+90=360

So the answer is 111 degrees

Correct me if im wrong! 

r =  2 * area  / perimeter

semi-perimeter,  S    =   (12 + 14 + 16)   / 2   =  21

Area  =  sqrt  [ 21  ( 21 - 16) (21 - 14) ( 21 -12) ]  =

sqrt [ 21 * 5 * 7 * 9 ]   =  21sqrt (15)

r  =  2 * 21 sqrt (15)  / 42  =   sqrt (15)

If we draw a line starting from B that is orthogonal to BC going down to AE at point Fand a line that starts at C that is orthogonal to BC as well but only goes to the same height as D at point G and connect the two points, we get two new 45-45-90 triangles, and AF+BC+GD=AE, and AF=BF, and BF=CG+DE=sqrt2+2, so sqrt2+2+2+sqrt2=2sqrt2+4=AE, so a=4, and b=2, so a+b=$\boxed{6}$

We only need to find the positive root......this will give us the  time

-16x^2  - 16x  + 60  =    0         

16x^2  + 16x  - 60  =  0       divide  through by  4

4x^2 + 4x - 15  =  0      factor

(2x - 3) ( 2x + 5)  =   0

Set  the  first factor =  0  and solve  for  x

2x - 3  =  0

2x =  3

x =  3/2 sec   = 1.5  sec

z=180-110=$\boxed{70^\circ}$

y=360-(93+80+70)=$\boxed{117^\circ}$

We see that this picture us constructed of squares and 45-45-90 triangles, where AB and FE are the sides of squares with length 1, as well as DC and GH which are sides of squares length 2, and HA and DE are the hypotenuses of 45-45-90 triangles, and CB and GF are hypotenuses of 45-45-90 triangles, so together, we have $(1\cdot1)+(1\cdot1)+(2\cdot2)+(2\cdot2)+(2\cdot2/2)+(2\cdot2/2)+(1\cdot1/2)+(1\cdot1/2)=\boxed{15}$

The white sector  of the circle =  (1/6) (10) =  10/6  =  5/3

So....the grey  area =   10 + 10  -  2 (5/3)  =  20 - 10/3  =  60/3  - 10/3  = 50 / 3

EH  = FG  ....so....

x - 3   = 4x  - 15

15  - 3  =  4x  -  x

12  =  3x

12 / 3   =  x   =  4

(-2)^2+y^2=4+y^2=x^2

(-1)^2+(y-4)^2=1+y^2+16-8y=x^2.  

4+y^2=y^2+1+16-8y

8y=13

y=$\boxed{\frac{13}8}$

Im getting 13/8.

26800 = 7300 ( 1+ .0775)ⁿ

26800/7300 =  1.0775ⁿ              Solve for n   ( you'll find it easiest to use LOGs)  

By unanimous consent......50%  it is    !!!!!

Note :   a "natural" number usually means  a "counring" number  ⇒ 1,2,3,4, ..... etc.

Jim is correct

Mikela is  not...  (-2 + -3 =  -5  which is not greater than  - 2  )

sry to ask again but i dont understand why you need to plus 1 in FV = 1000 ( 1 + .035/4)^20...

thank you sooooo much :)

Original arc length would be    x / 360   pi  2r  

 now multiply  x by 1.2   and   r by 1.25

1.2x / 360   pi  2 r * 1.25   =      1/2 * 1.25  x/360 pi 2r

                                                   = 1.5  *  original                        so 50% longer

Call  the original  arc length, S

Call the original radius, R

Call the orignal arc measure , T

So

S =  RT

And  when the arc measure  increases by 20%  and the  radius increases by 25%  we have

(1.20)R * ( 1.25) T  =

1.5  RT

So .....  the  arc length  increases  by   (1.5 -  1) * 100%  =  .5 * 100%  =   50%

by power of point, we have x^2=6x21=126

x=$\boxed{3\sqrt{14}}$

9(9+y)=81+9y=126

9y=45

y=$\boxed{5}$

Multiply numerator and denominator by   (1 - 7i)   to get

(9 + 37i) / 50       or      9/50 + 37i/50   

                                                  or     .18  +   .74 i

lets say the length radius is r, and the degree measure is d. we know that the length of the arc is 2$\pi$r$\cdot$d/360=d$\pi$r/180, so increasing the degree measure of the arc makes 2$\pi$r$\cdot$(6d/5)/360, and increasing the radius by 25%, we get 5$\pi$r/2$\cdot$(6d/5)/360 and we get:

$\frac{5\pi r \cdot \frac{6d}5}{720}=\frac{d\pi r}{120}$

$\frac{\frac{d\pi r}{180}}{\frac{d\pi r}{120}}=\frac32$

$\frac{3}{2} - 1=\frac{1}{2}=\boxed{50\%}$

\(\mathrm{Slope-Intercept\:form\:of}\:15x-6y=48:\quad y=\frac{5}{2}x-8\)