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x + 3y = kx,  3x + y = ky

Solve for x:

3x + y = ky

3x = ky - y

x = ky/3 - y/3

Subsitute into other equation:

x + 3y = kx

ky/3 - y/3 + 3y = kky/3 - ky/3

y(k/3 - 1/3 + 3) = y (kk/3 - k/3)

k/3 + 3 - 1/3 = kk/3 - k/3

k + 9 - 1 = kk - k

k + 8 = kk - k

8 = kk - 2k

k^2 - 2k - 8 = 0

Use Quadratic Formula.

x = 4, -2

The vertex of the parabola will be found half-ways between the focus and the directrix.

The vertex will be the point (4, 1.5).

A formula for the parabola is:  y - k  =  a(x - h)²  where (h, k) are the coordinates of the vertex.

For this situation:  y - 1.5  =  a(x - 4)².

Since the point  (8, 6)  is on the parabola:     6 - 1.5  =  a(8 - 4)²

                                                      --->      4.5  =  a(4)²     --->     4.5  =  16·a     --->     a  =  0.28125

The equation is:  y - 1.5  =  0.28125(x - 4)².

Multiplying out:   y - 1.5  =  0.28125(x² - 8x + 16)

                           y - 1.5  =  0.28125x² - 2.25x + 4.5

                                   y  =  0.28125x² - 2.25x + 6

You just finished telling us that you know all about polynomials now.

Let the hourly rate of the first mechnic =F
Let the hourly rate of the second mechanic=S

F + S = 160.........................(1)
20F + 5S =2,375..................(2), solve for F, S

Use substitution or elimination to get:
F =$105 - hourly rate of the first mechanic.
S =$55 - hourly rate of the second mechanic.

For k = 4, all values of x and y such that x = y work.  For all other values of k, the only solution is x = 0 and y = 0, so the unique value of k is 4.

Really? I never seen anyone call us cheaters. I'll be on the lookout then, thanks DragonLord!

Ralph is 10 years younger than 3 times Rick's age. That means Rick is younger than Ralph.

Let's add Ralphs age by 10 to find out 3 * Rick's age: 53 + 10 = 63.

63 is  3 * Rick's age. Divide by 3 = Rick's age.

63 / 3 = 21.

Rick is 21 years old.

I don't think I offended anyone but you, lol

Thanks for the A2A. I'm going to explain this very briefly. 

61% is 61/100 in fraction form. Now, we can see the probability of just 1 case. 

"find the probability that exactly 3 of them use their smartphones in meetings or classes". We can do \(\frac{61^3*39^5}{{100}^{8}}\) and get approximately .002. 

However, there are many cases so we will have to put in values for the combination formula.

After you substitute, you will get \(\frac{8!}{5!*3!}\). That is a total of 56 cases.

Multiply .00204 and 56. The answer in decimal form is  .11424.

In percent form, it is 11.424%. 

Cheers!

Same way to make a profit in any endeavor:  buy low, sell high. 

_.

The diagonals AC and BD of the convex quadrilateral ABCD with area 28 intersect each other at point O. The lines which are parallel to the diagonal AC come through the midpoints of segments BO and DO . Find the area of the portion of the quadrilateral that is between those lines.

Greetings,

I've written down your problem and solved it as understandable as possible.

See the images below.

Hope it's understandable.

Regards

It is 4 km.

1 - ∑[4^-n, n, 1, ∞ ] = 1 / 3

2 - ∑[ (3n - 2) / (3^n), n, 1, ∞ ] = 5 / 4

sorry, but thats wrong...

It is on your computer keyboard, generally above 1  !!!!!!!!.

Follow the same procedure of your first question and should get:

x = 45   and y = 60   and z = 75

Let Rick's age =A

Ralph's age     =3A - 10

   53                  =3A - 10

3A = 63

A =63 / 3 = 21 years - Rick's age.

There are 900 palindromes.  There are C(6,2) = 15 ways to choose two digits to swap, so there are 15*900 = 13500 swappy numbers.

Are you the same person that posted this question yesterday? If yes, did you look under that post?

See answers #1 and #2 here:  https://web2.0calc.com/questions/counting-probability

Look under your first post, some 15 questions below this one.

When Adam finishes, the gap between Ben and Charles is 4 km.  Also, the length of the race is 24 km, so the gap widens to 24/8*4 = 12 km by the end of the race.  So Charles is behind by 12 km when Ben finishes.