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     1/3 + 2/9  =  3/9 + 2/9  =  5/9    

     So 4/9 of the beans are orange.    

     (4/9)(total)  =  312     

     total  =  (9/4)(312)  =  702    

_.   

Since AK is the altitude, so we can use the pythaogrean thereom. 

We have'

$AK = \sqrt { AC^2 - CK^2 } = \sqrt { 8^2 - 3^2 } = \sqrt { 55 } \\ AB = \sqrt { AK^2 + BK^2 } = \sqrt { 55 + 4 } = \sqrt { 59 }$

Thanks! :)

https://web2.0calc.com/questions/right-triangles_35

The denominator would be $5\cdot5=25$. The only ways that the product is less than 4 is when they choose (1, 1) (1, 2), (1, 3), (2, 1), and (3, 1), which is 5 cases. So, $25 - 5 = 20$, so the answer is $\frac{20}{25} = \frac{4}{5}$.

Interior angles of an   n-gon sum to   ( n-2) 180

exterior angles sum to 360 

so

(n-2)(180) / n  = 2 *  360 / n     solve for 'n'

180 - 360/n = 720 / n 

180 = 1080 / n 

180 n = 1080 

n = 6   sides 

See image 

Find sqrt (72 ) using Pythag theorem 

then using Pythag theorem again:

 ( r+ sqrt(72) ^2  = r^2 + r^2 

r^2 - 2 sqrt(72) - 72 = 0                        Now use Quadratic Formula   with a = 1     b = -2 sqrt (72)    and   c = - 72

               to find r = 20.4853 units 

Catherine rolls a standard 6-sided die six times. If the product of her rolls is 60 then how many different sequences of rolls could there have been? (The order of the rolls matters.)    

5 6 2 1 1 1     ₆P₆ = 720    

5 3 4 1 1 1     ₆P₆ = 720    

5 3 2 2 1 1     ₆P₆ = 720    

Add them up.   

By my reckoning, 2160 different sequences.     

_.   

(d-7)  = (g-7) * 9        Given     this can be re-arranged to   (d+56) / 9 = g 

(d-4)  = ( g-4) * 6       Given       Solve this system for   g       sub in the underlined equation (above) for g  

d-4 = ( d+56)/9  -4 ) * 6 

d-4 = (6d + 336 ) /9 -24 

d-4 = 2/3 d  + 13 1/3 

1/3 d = 17 1/3 

d = 52    y/o                              (and g = 12 y/o) 

This does not have an answer   using  28/9   or    2  8/9    

The smallest integer Pearl wrote down is -40.

10 p * 10 min  /  1000 ft^2 =  1 p m / 100 ft^2     <====  it takes   1 people min to paint 100 ft^2 

Now with 5 people 

    1000 ft^2 *  ( 1 p m  / 100 ft^2 )   =  100  p m   

                then   100 p m / 5 p  = 20 min 

( if you halve the number of people ....you double the amount of time required) 

100 ft^2  / ( 10 m * 10 p) =    1 ft^2 / 1 m - p          each square foot requires 1  people  minute 

100 ft^2    * 1 p m / ft^2    /  15 p   =  6  2/3 minutes  

 How many miles is the distance from Penthaven to Jackson?

$s=vt\\ 2s=v\cdot 3.6h\\ 3.75h= \frac{s}{v+w}+\frac{s}{v-w}\\ 7.5h=\frac{2s}{v+w}+\frac{2s}{v-w}=\frac{v\cdot 3.6h}{v+30miles/h}+\frac{v\cdot 3.6h}{v-30miles/h}\\ 7.5=\frac{3.6v}{v+30miles/h}+\frac{3.6v}{v-30miles/h}\\ \color{blue}v=150\ miles/h$

$2s=v\cdot 3.6h\\ s=\frac{v\cdot 3.6h}{2}=150miles/h\cdot1.6h\\ \color{blue}s=240\ miles$

240 miles is the distance from Penthaven to Jackson.

  !

I = V/R

$R_1=\frac{V_1}{I_1}=\frac{500V}{0.025A}=\color{blue}20k\Omega$

I halve the length of the wire, hence the resistance $R_1$.

$I_2=\frac{V_2}{R_2}=\frac{175V}{10k\Omega}=\color{blue}17.5mA$

I apply the voltage$V_2$ to both halves of the wire, thus halving the resistance $R_1$.

$I_3=\frac{V_2}{R_3}=\frac{175V}{5k\Omega}=\color{blue}35mA$ in both wire halves together.

 !

Find the two y values corresponding to x = 4 

4^2 + y^2 = 25 

    y = + - 3           so now find the distance between   (4,3) and ( 4, -3) 

                  distance = 6      between A and B 

The 'x' coefficient   will be the result of   1 * (-23x)   + 14 * x   =   -23x + 14x  =   - 9x  

so the coefficient in question will be -9 

First, let's find the total number of cases we work through. We can actually just list them all out. 

We have

$(HHT,HTH,HTT, THH, THT, TTH, TTT)$

The only possibility where a head doesn't appear is TTT. That means that

$(HHT,HTH,HTT, THH, THT, TTH)$ all work. 

Since probability is essentially $\text{Probability} = \frac{\text{# of Favorable Outcomes}}{\text{Total Outcomes}}$

So we just have $\frac{6}{7}$

So 6/7 is our answer. 

Thanks! :)

We can use the great pythaogrean theorem to our advantage. 

First off, let's set some variables. 

Let's let BC = a and AC = b. 

Now, using pythagoras, let's right some equations. We have

$\overline{AM} = \sqrt{b^2 + \left(\dfrac a2\right)^2} = \dfrac12 \sqrt{4b^2 + a^2}$ since AM is the median and we also have

$\overline{BN} = \sqrt{a^2+ \left(\dfrac{b}{2}\right)^2} = \dfrac12 \sqrt{4a^2 + b^2}$ since BN is also the median of the triangle. 

Now, we're trying to isolate a^2+b^2 because that's how we find the hypotenuse of the triangle. 

Thus, we now have

$4b^2 + a^2 = 2^2 \\4a^2 + b^2 = 2^2$

Now, adding the two equations up with each other and dividing both sides by 5, we find that

$a^2 + b^2 = \dfrac{4+4}{5} = 8/5$

So the hypotenuse is that square rooted, which is $\sqrt{\frac{8}{5}}$

So $\sqrt{\frac{8}{5}}$ is our final answer. 

Thanks! :)

x * y * 42 = 630 

x*y = 15 

x and y   are   3 and 5    or    15 and 1