hectictar

√5  is an irrational number

y tan 33  =  (y + 855) tan 25

                                                           Distribute  tan 25  to both terms in parenthesees.

y tan 33  =  y tan 25 + 855 tan 25

                                                           Subtract  y tan 25  from both sides of the equation.

y tan 33 - y tan 25   =   855 tan 25

                                                           Factor  y  out of both terms on the left side.

y( tan 33 - tan 25 )   =   855 tan 25

                                                           Divide both sides of the equation by  ( tan 33 - tan 25 ) .

y   =   855 tan 25 / ( tan 33 - tan 25 )

The graph shown is the graph of  y = 4/x  shifted  6  units left and  2  units up, so it has the equation

y  =  4 / (x + 6) + 2

y  varies directly with  x  and inversely with  z .

y  =  kx / z

                            And  y = 18  when  x = 15  and  z = 5 .

18  =  k(15) / 5

                            Multiply both sides by  5 .

90  =  k(15)

                            Divide both sides by  15 .

6  =  k                  So...

y  =  6x / z

Find  y  when  x = 21  and  z = 7 .

y  =  6(21) / 7

y  =  126 / 7

y  =  18

1)      w  and  t  vary inversely, so...

t  =  k / w

                               t = 1/5   when  w = 4  ,   so...

1/5  =  k / 4

                               Multiply both sides of the equation by  4

4 * (1/5)  =  k

4/5  =  k                  Now we can use this value of  k in the equation  w = k / t

t  =  (4/5) / w

                               Simplify.

t  =  (4/5) * (1/w)

t  =  4 / (5w)            To find t  when  w = 9 ,  plug in  9  for  w  in this equation.

t  =  4 / (5(9))

t  =  4 / 45

width   =   4  yards

length   =   5\(\frac23\)  yards

area   =   width  *  length

                                                      Plug in the given values for width and length.

area   =   4 yards  *  5\(\frac23\)  yards

area  =   (4 * 5\(\frac23\))  square yards

                                                      Convert  5\(\frac23\)  to  \(\frac{17}{3}\)

area  =   (4 * \(\frac{17}3\))  square yards

                                                      Multiply.

area   =   \(\frac{68}{3}\)   square yards

                                                      Convert  \(\frac{68}3\)  to  22\(\frac{2}{3}\)

area   =   22\(\frac23\)  square yards

1)

P(x)   =   55,000√[x - 1945]

In which year was the population 275,000  ?

What is the value of  x  when  P(x)  =  275,000   ?

275,000   =   55,000√[x - 1945]

                                                     Divide both sides by  55,000 .

5   =   √[x - 1945]

                                             Square both sides of the equation.

25   =   x - 1945

                                             Add  1945  to both sides.

1970   =   x

In the year  1970  , the population was  275,000 .

2)

\(f(x)\,=\,\sqrt[3]{x-2}\)

                                       Instead of  f(x)  let's write  y .

\(y\,=\,\sqrt[3]{x-2}\)

                                       Now let's solve this equation for  x . Raise both sides to the power of  3

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

                                       Add  2  to both sides.

\(y^3+2\,=\,x \\~\\ x\,=\,y^3+2\)

                                       So the inverse function is...

\(f^{-1}(x)\,=\,x^3+2\)

After the last round, A has 20 counters, B has 10, and C has 9 . So...

(p + q + r) * the num of rounds  =  20 + 10 + 9   =   39

p + q + r   =   39 / the num of rounds

Since  p + q + r  must be an integer, the num of rounds must be  3,  13,  or  39 .

If the num of rounds = 13 ,  p + q + r  =  3  , but no 3 different positive integers add to 3 .

If the num of rounds = 39 ,  p + q + r  =  1  , but no 3 different positive integers add to 1 .

So the num of rounds must be 3 , and  p + q + r = 13

By trial and error, I think I found 3 different integers that work:  1, 4, and 8

sum of counters earned in 3 rounds for player A   =   8 + 8 + 4  =  20
sum of counters earned in 3 rounds for player B   =   1 + 1 + 8  =  10
sum of counters earned in 3 rounds for player C   =   4 + 4 + 1  =  9

q  =  the middle number  =  4

On the first round, player C earned  4  counters.

I just got lucky by guessing those numbers for p, q, and r.....Maybe there is a way to do it without guessing!

1)  " Triangle PQR~Triangle PUT, not similar. "

This says, "triangle PQR is similar to triangle PUT, not similar."

If they are not similar then we can't make a similarity statement!

But these triangles actually are similar...

∠PQR  ≅  ∠PUT   because they both are marked by a red line, and

∠QPR  ≅  ∠UPT   because they are vertical angles.

Since two of the angles are the same, the triangles are similar.

△PQR ~ △PUT

2)

∠VTU  ≅  ∠DTE  because they are vertical angles.

If these triangles are similar, then this must be true of the sides adjacent to the angle we know is shared by both triangles:

\(\frac{\text{shorter side of first triangle}}{\text{shorter side of second triangle}}\,=\,\frac{\text{longer side of first triangle}}{\text{longer side of second triangle}} \\~\\ \frac{19}{63}\,=\,\frac{27}{98} \)

Since this is not true, the triangles are not similar.

3)

If the triangles are similar, then this must be true:

\(\frac{32}{56}\,=\,\frac{40}{70}\,=\,\frac{48}{84} \\~\\ \frac47\,=\,\frac47\,=\,\frac47\)

Since this is true, the triangles are similar.

△EDC ~ △JKL