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We see that that some of the points on line l are (0, 5) and (3, 0) and some of the points that are on line m are (0, 2) and (7, 0).

For line l, for every increase of 3 in the x direction, there is a decrease of 5 in the y direction. Since we want y = 15, we will increase y two times to get to (0, 15). This will cause x to decrease by 3 two times as well, so we'll get (-6, 15) for line l. For line m, we want to increase 13 units upwards, and every 2 units up causes 7 units left. Therefore, (13)(7/2) = 91/2. 

Therefore, the coordinates of the point on line m is (-91/2, 15). The difference in the x-coordinates are -6-(-91/2) = 79/2

Have you worked out what the vertices are yet?  

Hi lovely

\(\sqrt[5]{x}^{\frac{5}{6}}*\sqrt{x}\\ =(x^{\frac{1}{5}})^{\frac{5}{6}}*x^{\frac{1}{2}}\\ =(x^{\frac{1}{5}*\frac{5}{6}})*x^{\frac{1}{2}}\\ =x^{\frac{1}{6}}*x^{\frac{3}{6}}\\ =x^{\frac{4}{6}}\\ =x^{\frac{2}{3}}\\ =\sqrt[3]{x^2}= (\sqrt[3]{x})^2\)

Solve \(\large \sqrt{x+14-8\sqrt{x-2}} + \sqrt{x+23-10\sqrt{x-2}} = 3\)

Substitute: \(\begin{array}{|rcll|} \hline \mathbf{ y } &=& \mathbf{\sqrt{x-2}} \qquad \text{or} \qquad y^2 &=& x-2 \\ \mathbf{ x } &=& \mathbf{y^2+2} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \sqrt{x+14-8\sqrt{x-2}} + \sqrt{x+23-10\sqrt{x-2}} &=& 3 \\\\ \sqrt{y^2+2+14-8y} + \sqrt{y^2+2+23-10y} &=& 3 \\\\ \sqrt{y^2-8y+16} + \sqrt{y^2-10y+25} &=& 3 \\\\ && \boxed{ y^2-8y+16 = (y-4)^2=(4-y)^2} \\ && \boxed{ y^2-10y+25 = (y-5)^2=(5-y)^2} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline \text{Case }1: \\ & y^2-8y+16 = (y-4)^2 \\ & y^2-10y+25 = (y-5)^2 \\ \hline & \sqrt{(y-4)^2} + \sqrt{(y-5)^2 } &=& 3 \\\\ & y-4 + y-5 &=& 3 \\\\ & 2y-9 &=& 3 \\\\ & 2y &=& 9+3 \\\\ & 2y &=& 12 \\\\ & \mathbf{y} &=& \mathbf{6} \\ \hline \end{array} \begin{array}{|lrcll|} \hline \text{Case }2: \\ & y^2-8y+16 = (y-4)^2 \\ & y^2-10y+25 = (5-y)^2 \\ \hline & \sqrt{(y-4)^2} + \sqrt{(5-y)^2 } &=& 3 \\\\ & y-4 + 5-y &=& 3 \\\\ & \mathbf{1} &\ne& \mathbf{3} \qquad \text{no solution!} \\ \hline \end{array}\\ \begin{array}{|lrcll|} \hline \text{Case }3: \\ & y^2-8y+16 = (4-y)^2 \\ & y^2-10y+25 = (5-y)^2 \\ \hline & \sqrt{(4-y)^2} + \sqrt{(5-y)^2 } &=& 3 \\\\ & 4-y + 5-y &=& 3 \\\\ & -2y+9 &=& 3 \\\\ & 2y &=& 9-3 \\\\ & 2y &=& 6 \\\\ & \mathbf{y} &=& \mathbf{3} \\ \hline \end{array} \begin{array}{|lrcll|} \hline \text{Case }4: \\ & y^2-8y+16 = (4-y)^2 \\ & y^2-10y+25 = (y-5)^2 \\ \hline & \sqrt{(4-y)^2} + \sqrt{(y-5)^2 } &=& 3 \\\\ & 4-y + y-5 &=& 3 \\\\ & \mathbf{-1} &\ne& \mathbf{3} \qquad \text{no solution!} \\ \hline \end{array}\\\)

\(\begin{array}{|rcll|} \hline \text{First solution}: \\ y= 6 \\ \hline x &=& y^2+2 \\ x &=& 6^2+2 \\ \mathbf{x} &=& \mathbf{38} \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{Second solution}: \\ y= 3 \\ \hline x &=& y^2+2 \\ x &=& 3^2+2 \\ \mathbf{x} &=& \mathbf{11} \\ \hline \end{array}\)

The solid is a couple of cones that share the same base.

My diagram may help you in solving this geometric problem.

As follows:

Solve this system.

The hypotenuse of an isosceles right triangle is 4sqrt(2) units. How many square units are in the area of the triangle?

A = ( 4√ 2 )² / 4 = 8 u²

Let Y, X, and W be three consecutive vertices of a regular 18-gon. A regular heptagon is constructed on Line XY,  with a vertex Z next to X. Find angle YZW, in degrees.

Angle YZW = 80 degrees  

The sum of a two digit number and the number obtained by interchanging the digits is 154.If the tens digit number is 2 more than units digit number find the original number.

Hi

If anyone can answer please, what is the concept of \(\lambda \)  in sums that Alan used called?

When a quadrilateral is inscribed in a circle, the sum of the opposite angles is 180º!!!

Angle  a = 110º / 2 = 55º

Angle  b = 180º - 55º = 125º  

Your question is not displayed well (you have not bothered to delete meaningless dollar signs)

And

You go out of your way to irritate established answerers.

So why would you expect to get quick, high quality, answers ?

Sorry! Well, I tried :(

d / 3      =   d/5   + 2/3      solve for  d       ( 2/3  is 40 minutes )

d  = 3/5 d + 2

2/5 d = 2

d = 5  miles

Thank you!

At a concession​ stand,three hot dog(s) and two hamburger(s) cost ​$8.25​;two hot dog(s) and three hamburger(s) cost ​$8.00. Find the cost of one hot dog and the cost of one hamburger. What is the cost of one hot dog​?  

                                                              3 Dogs + 2 Burgers = 8.25    (1)  

                                                              2 Dogs + 3 Burgers = 8.00    (2)  

Multiply equation (1) by 2                       6D + 4B = 16.50                    (3)  

Multiply equation (2) by 3                       6D + 9B = 24.00                    (4)  

Subtract equation (3) from (4)                        5B  =  7.50  

Divide both sides by 5                                       B  =  1.50   One Burger costs $1.50  

Substitute 1.50 for B in (1)                      3D + (2)(1.50)  =  8.25  

                                                                3D + 3.00         =  8.25  

Subtract 3.00 from both sides                                    3D  =  5.25  

Divide both sides by 3                                                  D  =  1.75   One Dog costs $1.75  

_.

Multiply first equation by 3 and re-arrange   to  x = 3-6y     Substittute this value of x into the second equation

y = 2/3 ( 3-6y) - 4      solve this equation for y    then use this value of y in one of the equations to find x

G =   s*52 / 365                grandson age in years

A =   s*52/365   * 12         adult's age  (what the question is asking )

s = son's age                    summed = 120

s*52/365   +  s   +  s*52/365  * 12  =  120

solve for s = 42 y/o

then  G = 6 y/o

A = 42*52/365  * 12 =  ~72 y/o

Okay.... thanks

13 because 5 minerals multiplied by 3 roots is 15. 1 of the minerals is not compatible with 2 of the roots. So all you have to do is subtract 2 from 15.

=13

1009/1000 = 100.9%

ok, so you think 

P(n)^2 =n    means

[ P(n) ]^2 =n

Yes, that is probably what is meant.

G = years (Grandpa)
G=Months(Grandson)=G/12
G/12*365 /(365/7)=Son's age in years.
G + G/12 +[G/12*365 /(365/7)] =120, solve for G
G=72 years - Grandpa's age
72 months / 12 = 6 Years - Grandson's age. So that:
120 - 72 - 6 =42 years - son's age. So that:
6 years x 365 days = 42 years x (365/7) weeks
2,190 days - grandson's age  = 2,190 weeks - son's age

I solved this using coordinates:

X = (1,0)

Y = (0.924328, 0.471208)

Z = (1.222783, -1.147542)

W = (0.924328, -0.471208)

==> angle BCD = 75 degrees