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Your question is worded improperly.....

I will assume you want to know sis age when xing becomes half his father's age at the time  ...in the future

15+x = 1/2 (46+x)         where x is th years from now when xing reaches 1/2 father's age

.5x - 8

x = 16      more years        8+16 = sis age = 24 y/o

\(\frac{\sqrt{32}}{\sqrt{16} - \sqrt{2}} = \frac{2 + 6 \sqrt{2}}{7}\)

 A+ B + C+ D = 2 + 6 + 2 + 7 = 17

2x - 4y = -7.

First, we need to find the side lengths. Since the side length of a square is the square root of the area, we now know that the two side lengths are

\(\sqrt {32} \text{ and } \sqrt{63}\)

The ratio would now be \(\frac{\sqrt{32}}{\sqrt{63}}\).

Simplifying the square roots, we get \(\frac{4\sqrt 2}{3\sqrt 7}\).

From here, we need to multiply the top and bottom by \(\sqrt 7\) to rationalize the denominator. So, we get

\(\frac{4\sqrt {14}}{21}\)

Now, all you have to do is add a, b, and c. This will give us

\(4 + 14 + 21 = 39\)

You are close! 

Let's set up a proportion, where x is the number of men doing the job in 15 days, and y is the number of days 10 men need to do the job. 

\(\frac{x}{15} = \frac{10}{y}\)

Cross multiplying will give us xy = 150.

Since we need to find y, we now need to find x in terms of y. This will give us x = 150/y.

Now, our new proportion is 

\(\frac{\frac{150}{y}}{15} = \frac{10}{y}\)

Now, we can either form a system of equations, OR, we can directly see that when y is 15, all the euqations work out. 

Therefore, you will need 15 days for 10 men to do the job. 

I would start by drawing a probability tree.  Do you know how to draw one?

Fred draws three cards from a standard deck of 52 cards. First he drew a Jack, then a Queen, then a King.

What is the probability of this happening, in this exact order? Assume the drawing is done without replacement.

prob = (4/52)*(4/51)*(4/50)

Two 4 x 4 squares intersect at right angles, bisecting their intersecting sides, as shown. The circle's diameter is the segment between the two points of intersection. What is the area of the shaded region created by removing the circle from the square?

Hello Guest!

\(A=2\cdot (4\cdot 4)-2^2=28\)

\(d=\sqrt{2^2+2^2}=\sqrt{8} \\d=2\cdot \sqrt{2}\\ r=\sqrt{2}\)

\(A_{shaded}=A-\pi r^2=28-\pi \cdot 2=28-6.28\)

\(A_{shaded}=21.72\)

  !

I prove it

\(x+y+z=2\\ xy-z=-3\\ x+y+z^2=4 \)

\(x=-1\\ y=1\\ z=2\)

\(-1+1+2=2\\ -1\cdot 1-2=-3\\ -1+1+2^2=4 \)

q.e.d.

Mistake!
The question asks for a unique solution. The equations are solved by both x = -1, y = 1, z = 2 and x = 1, y = -1, z = 2.
The correction follows.

  !

Area ≈ 174.2 squared units 

write a system of 3 equations and 3 variables that has the unique solution of (-1,1,2).

Hello Guest!

\(x+y+z=2\)

\(x y-z=-3\)

\(x+y+z^2=4\)

  !

Find the surface area.

Hello Guest!

\(A=2\times(\frac{12+8}{2}\times 3+8\times 10)+5\times 10+3\times 10\)

\(A=300\)

  !

four ways to get to 8 in just using the number 2,3 and 4

1)     2⁰ + 3 + 4 = 8

2)     2 * 3 + √4 = 8

3)     ( 2 * 3 * 4 ) / 3 = 8

4)      3² -  (√4) / 2  = 8    

Sum up the sequence first using: n * (n+1) / 2 formula:[80 * 81] / 2 =3,240

Then factor 3240 = 2^3 * 3^4 * 5

As you can see, the largest Prime factor = 5

1)    The angle bisector theorem:    CY / AY = BC / AB

                                                    4 / 6 = BC / 8

                                                     BC = 5.333333

2)    Using a Law of cosines calculate angles C  and  B

3)    Use the Law of sines to calculate BZ

Geometry is fun.

If you let x go infinity, then you get 0 = A + B.

If you let x = 0, then you get -7/2 = A/(-2) + B.

The solution to this system is then A = 7/3, B = -7/3, so (A,B) = (7/3,-7/3).

The bounded are is 14.

The probability is 1/13*1/13*1/13 = 1/2197.

Just gather all of the 'like' terms:

(x⁴ +7x⁴) + ( 3x³ - 2x³) + ( - 5x² + x²)  + (-x + x)   + 1       Can you take it from here?  

You don't need the law of sines, you need this:

15m / tan(55º) ≈ 10.5 m  

(x-0)^2    + y^2 - 6y   = 16                       complete the square on x and y

(x-0)^2         + (y-3)^2    = 16+9              does that help?

                                                  Do you remember the standard form of a circle equation?

Thanks very much for the explanation.

I do not remember seeing such a question done like this before.

You have taught me, I hope I retain the knowledge.