hectictar

average speed for the whole trip  =  total distance / total time

total distance  =  120 miles + 150 miles  =  270 miles

total time  =  2 hours + 1 hour + 3 hours  =  6 hours

average speed  =  270 miles / 6 hours  =  45 miles / 1 hour

So her average speed is 45 miles per hour.  

A(t)  =  3 - 2t² + 4^t

A(2)  =  3 - 2(2)² + 4²

A(1)  =  3 - 2(1)² + 4¹

A(2) - A(1)  =  [ 3 - 2(2)² + 4² ] - [ 3 - 2(1)² + 4¹ ]

A(2) - A(1)  =  [ 3 - 2(4) + 16 ] - [ 3 - 2(1) + 4 ]

A(2) - A(1)  =  [ 3 - 8 + 16 ] - [ 3 - 2 + 4 ]

A(2) - A(1)  =  [ 11 ] - [ 5 ]

A(2) - A(1)  =    6

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f( √[x + 1] )  =  1/x      for  x ≥ -1  and  x ≠ 0  

We want to find  f(2)  , so we need an x value that makes

√[ x + 1]  =  2      square both sides

x + 1  =  4           subtract  1  from both sides

x  =  3                 This is a valid  x  to plug in.

So....

f( √[3 + 1] )  =  1/3

f( 2 )  =  1/3                 

y  =  5x - 4

a.  The x-intercept is the value of  x  when  y = 0 , so plug in  0  for  y  and solve for  x .

0  =  5x - 4      Add  4  to both sides of the equation.

4  =  5x           Divide both sides by  5 .

4/5  =  x

The x-intercept is  4/5  , or the point  (4/5, 0)  .

b.  Now we need to find the y-intercept. So plug in  0  for  x  and solve for  y .

y  =  5(0) - 4

y  =  -4

The y-intercept is  -4  , which is the point  (0, -4) .

To graph this, plot the points  (4/5, 0)  and  (0, -4)  and draw a line through them.

\(27^{-\frac{1}{2}} \\~\\ =\,\frac1{27^{\frac12}} \\~\\ =\,\frac1{\sqrt{27}} \\~\\ =\,\frac{1}{\sqrt{3*3*3}} \\~\\ =\,\frac1{3\sqrt3} \\~\\ =\,\frac{\sqrt3}{9}\)

\(3\frac18-1\frac78 \\~\\ =\,(3+\frac18)-(1+\frac78) \\~\\ =\,3+\frac18-1-\frac78 \\~\\ =\,3-1+\frac18-\frac78 \\~\\ =\,2-\frac68 \\~\\ =\,1+\frac88-\frac68 \\~\\ =\,1+\frac28 \\~\\ =\,1\frac14\)

a - 4(a + b)            Distribute the  -4  to each term in the parenthesees.

=   a  -  4a  -  4b         Combine  a  and  -4a  to get  -3a  .

=  -3a - 4b

2x( 3 - x )

=   2x * 3   +   2x * -x

=   2 * 3 * x   +   2 * x * x * -1

=   6x   +   - 2x²

=   6x - 2x²

rate   =   distance / time   =   5.5 miles / 0.5 hours   =   11 miles / hour

distance   =   rate * time   =   11 miles / hour  *  t  hours   =   11t miles

distance  =  11t miles

To find how far Brandon bikes for each time interval, plug in whatever time they give you in hours.

If you draw a picture of his path, it looks like this..

It makes a right triangle, and side AC is the width of the river. In every right triangle,

tan θ  =  opposite / adjacent

tan 54°  =  AC / 70                   Multiply both sides by  70 .

70 tan 54°  =  AC

96 feet  ≈  AC

1)     Triangle PQR  is similar to triangle PST , so

PQ / PS  =  PR / PT          And  PQ  =  8 ,  PS  =  8 + 6 ,  and  PT  =  10.5 .

8 / [8 + 6]  =  PR / 10.5     Multiply both sides of this equation by  10.5 .

10.5 * 8 / [8 + 6]  =  PR     Simplify.

6  =  PR

2)     Notice that we have two parallel lines cut by a transversal, and their alternate angles are congruent. So we can be sure that  triangle OMN  is similar to triangle OQP  , which means

OQ / OM  =  OP / ON      And  OQ  =  6 ,  OM  =  9 , and  ON  =  25 - OP

6 / 9  =  OP / [25 - OP]    Solve for  OP .

10  =  OP

3)     Angle CAB  is a part of both triangles, and both triangles have a right angle.

So  triangle ABC  is similar to  triangle ADE  by angle-angle similarity.

AB / AC  =  AD / AE      And  AB  =  11 + 10  , AC  =  35 ,  and AE  =  11

[11 + 10] / 35  =  AD / 11

11 * 21/35  =  AD

4)     Notice that angle DEF and angle CEB are vertical angles, angle BFD and angle FBC are alternate, and the other two are also alternate. So  triangle DEF  is similar to  triangle CEB .

DF / DE  =  CB / CE     Since  CD and CB are sides of the same square, CD = CB = 2 + 5

DF / 2  =  [2 + 5] / 5      I'll let you finish solving this for  DF .

5)     Triangle MNO  is similar to  triangle QPO , so  ∠MNO  =  ∠QPO  =  60°

There are  180°  in every triangle. ∠RNO  =  ∠PQO  =  180° - 60° - 66°  =  54°

∠MNR  =  ∠MNO - ∠RNO  =  60° - 54°  =  6°