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Using the midpoint formula, we have :

[ 6 + x ] / 2   = 1       and     [ 0 + y ] / 2  = 10

Multiply both sides by 2

6 + x   = 2                                                0 + y = 20

subtract 6 from both sides

x = -4                                                             y = 20

So.......the other endpoint is  ( -4, 20).......just as EP found    !!!!

x = 2

x = 0

x = 2

because you would add the negative 1 to the opposite side and then multiply by 2.

Yep....more songs is always a good idea....!!!!

From the endpoint to the midpoin  x  goes from  6 ---->  1   (or -5)     move x another -5 to get the other endpoint  x = -4

                                                       y                    0----> 10  (or +10)             y              10                                         y= 20

Other endpoint is thus   (-4, 20)

logx143=0.96

x^.96 = 143

x = 143^(1/0.96)

x =175.85, because

175.85^0.96 = 143

p + 1/6 = 6/11

p = 6/11 - 1/6

p = 36/66 - 11/66 = 25/66

25 minute : 4 hours x 60 minute/hr

25 : 240

5:48

If there is no air friction.....when the rock comes down to its launch point it will be going the same speed as when it was launched, so it will be traveling downward at the same speed as the first rock was thrown (down) and will have the same energy at impact:   #2   K joules

even, if you multiply a number by 10, all you are doing is adding a 0 at the end. any number that ends in 0 is even

You cannot graph it as it because there is no y

logx143=0.96

\(log_x143=0.96\\ \frac{log143}{logx}=0.96\\ logx=log(143)/0.96\\ 10^{logx}=10^{log(143)/0.96}\\ x=10^{log(143)/0.96}\\\)

10^(log(143)/0.96) = 175.84972994163184

\(x\approx 175.84972994163184\)

check

log(143,175.84972994163184) = 0.9599999999999999

you are not goint to get any closer than that :)

Come on rarinstraw, we are not here to do all your homework for you.    

Try them with some numbers ans see what happens. 59 looks pretty easy.

63 is easy

If you look at the formulary on this site the log rules are there.

You have a right triangle with hypotenuse = 61 m  (given)

One leg is  11 m (given) 

Find the OTHER leg and add the pedestal heiht of 5m to it to find the highest point

a^2 + b^2 = c^2

a^2 + 11^2 = 61^2

a^2 = 61^2   -  11^2

a^2 = 3600

a =60

60 +5 = 65 m  is how high the monument is at the top.

it is 1

an airplane is 33 miles due south and 56  miles due west of its destination airport. How far is the plane from the destination airport

Quadratic graphs !!!

how do you do y=2xsq-x-3 on a table of values

\(y=2x^2-x-3\)

x    f(x)
-10    207,0
-9,5    187,0
-9    168,0
-8,5    150,0
-8    133,0
-7,5    117,0
-7    102,0
-6,5    88,00
-6    75,00
-5,5    63,00
-5    52,00
-4,5    42,00
-4    33,00
-3,5    25,00
-3    18,00
-2,5    12,00
-2    7,00
-1,5    3,00
-1    0,00
-0,5    -2,00
0    -3,00
0,5    -3,00
1    -2,00
1,5    0,00
2    3,00

asinus :- )    !

a suitcase measures 24 inchers long and 18 inches high. What is the diagonal lenght of the suitcase

Let x = #students

60% *  x  = 18

.6x = 18     divide both sides by  .6

x = 18/.6

x = 30 students

(5x-1/4)=25

\(5x-\frac{1}{4}=25\)

\(5x=25\frac{1}{4}\)

\(x=5\frac{1}{20}\) = 5.05

  !

If you use the common core method, you would change the 9 to 1/9, that makes it into a multiplication problem, 1/2 times 1/9 is 1/18

write an equation of the line that passes through the points

(3,8) (-2,8)

\(m=\frac{y_2-y_1}{x_2-x_1}\)

\(m=\frac{8-8}{-2-3}=\frac{0}{-5}=0\)

\(m=0\)

\(y_1=mx_1-b\)

\(b=mx_1-y_1\)

\(b=0\times 3- 8\)

\(b=-8\)

\(y=mx+b\) 

\(y=0\times x-8\)

\(y=-8 \)    \(is \ the \ equation.\)

  !

\(Simpler \ solution: y_1=y_2=-8\)

\(y=-8\)  \(is \ the \ equation.\)

Try this:
sin^-1(-1/2)

i dont know

you eat it

'solute'

Make the number = x

9x -4 = 176

9x= 180

x = 20