hectictar

Here is my guess...

From  E  to  F  there are  4  different paths:

South, East, East, East

East, South, East, East

East, East, South, East

East, East, East, South

From  F  to  G  there are  10  different paths:

South, South, South, East, East

South, South, East, South, East

South, South, East, East, South

South, East, South, South, East

South, East, South, East, South

South, East, East, South, South

East, South, South, South, East

East, South, South, East, South

East, South, East, South, South

East, East, South, South, South

4 * 10  =  40

I think there are  40  different 9-step paths from  E  to  G  which pass through  F .

If this is right, maybe someone else can give a better explanation!

Thanks X²

18)

90 * the scale factor   =   100

the scale factor   =   100 / 90

the scale factor   =   10/9

8)

We can think of this as the point that divides the line segment in a 1:1 ratio.

x-coordinate   =   -5 + 1/(1 + 1) * (2 - -5)   =   -5 + 1/2 * 7   =   -5 + 7/2   =   -1.5

y-coordinate   =    3 - 1/(1 + 1) * (3 - -2)   =    3 - 1/2 * 5   =    3 - 5/2   =   0.5

The coordinates are  (-1.5, 5.5)

1.

∠ACB  ≅  ∠ECD   because they are vertical angles.

∠ABC  ≅  ∠EDC   because they are both right angles.

So by the AA similarity (postulate or theorem? I don't know..)  △ABC  ~  △EDC .

2.

AB / BC   =   ED / DC

AB / 79    =   57 / 23

3.

AB / 79   =   57 / 23

                                                   Multiply both sides of the equation by  79 .

79 * AB / 79   =   79 * 57 / 23

                                                   Simplify.

AB   =   79 * 57 / 23

AB   =   79 * 57 / 23

AB   =   4503 / 23

AB   ≈   196   feet

\(\quad\sqrt[3]{16}\,\times\,7\sqrt[3]{2} \\ =\\ \quad \sqrt[3]{16}\,\times\,7\,\times\,\sqrt[3]{2} \\ =\\ \quad \sqrt[3]{16}\,\times\,\sqrt[3]{2}\,\times\,7 \\ =\\ \quad \sqrt[3]{16\,\times\,2}\,\times\,7 \\ =\\ \quad \sqrt[3]{2\,\times\,2\,\times\,2\,\times2\,\times\,2}\,\times\,7 \\ =\\ \quad \sqrt[3]{2^3\,\times\,2^2}\,\times\,7 \\ =\\ \quad \sqrt[3]{2^3}\,\times\,\sqrt[3]{2^2}\,\times\,7 \\ =\\ \quad 2\,\times\,\sqrt[3]{4}\,\times\,7 \\ =\\ \quad 2\,\times\,7\,\times\,\sqrt[3]{4} \\ =\\ \quad 14\,\times\,\sqrt[3]{4} \\ =\\ \quad 14\sqrt[3]{4} \\ \)

\(4x^2 - 31x = 8\)

                                      Subtract  8  from both sides of the equation.

\(4x^2 - 31x - 8 = 0\)

Now we can solve this quadratic equation using the quadratic formula.

The quadratic formula says...

If     \(ax^2+bx+c=0\)    then    \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

So...

If    \(4x^2 - 31x - 8 = 0\)    then    \(x = {-(-31) \pm \sqrt{(-31)^2-4(4)(-8)} \over 2(4)}\)

\(x = {-(-31) \pm \sqrt{(-31)^2-4(4)(-8)} \over 2(4)} \\~\\ x = {31 \pm \sqrt{961+128} \over 8} \\~\\ x = {31 \pm \sqrt{1089} \over 8} \\~\\ x = {31 \pm 33 \over 8} \\~\\ \begin{array} \ x=\frac{31+33}{8}\qquad&\text{or}&\qquad x=\frac{31-33}{8} \\~\\ x=\frac{64}{8}&\text{or}&\qquad x=\frac{-2}{8} \\~\\ x=8&\text{or}&\qquad x=-\frac{1}{4} \end{array}\)

The problem says that   y  =  8x + 17  .

To find  y  when  x = 0  ,  plug in  0  for  x  and solve for  y .

When  x = 0 ,   y  =  8(0) + 17   =   0 + 17   =   17

When  x = 1 ,   y  =  8(1) + 17   =   8 + 17   =   25

When  x = 2 ,   y  =  8(2) + 17   =  16 + 17  =   33

When  x = 3 ,   y  =  8(3) + 17   =  ...can you finish this?

And this is how to fill in the table...

Can you finish the last two?

Reason

1.   Given

2.   Vertical angles are congruent

3.   AA Similarity Theorem

Both triangles have a 90° angle. In order for the triangles to be similar, the sides that form this angle must be proportional, by the SAS Similarity Theorem.

If  △PRQ ~ △MRN , then  PR / RQ  =  MR / RN

Is it true that   PR / RQ  =  MR / RN   ?

PR / RQ  \(\stackrel{?}{=}\)  MR / RN

                                       Plug in the lengths of PR, RQ, MR, and RN.

12 / 18  \(\stackrel{?}{=}\)  8 / 6

                                      Reduce  12/18  by  3 .

4 / 6   \(\stackrel{?}{=}\)   8 / 6

4 / 6   ≠   8 / 6

Therefore  △PRQ  is not similar to   △MRN