hectictar

Examples:

To enter:     $\frac12$  ,     type:     1/2

To enter:     $\frac{15}{16}$  ,     type:     15/16

Simplify first equation:

2x - 3y  =  -3.2 - 0.2x + 0.1y

                                                Add  0.2x  to both sides of the equation.

2.2x - 3y  =  -3.2 + 0.1y

                                                Add  3y  to both sides.

2.2x  = 3.1y  - 3.2

Simplify second equation:

x  =  0.6x - y + 8.8

                                      Add  y  to both sides of the equation.

x + y  =  0.6x + 8.8

                                      Subtract  x  from both sides.

y  =  -0.4x + 8.8

Substitute  -0.4x + 8.8  in for  y  in the first equation and solve for  x .

2.2x  =  3.1( -0.4x + 8.8 ) - 3.2

                                                     Distribute  3.1  to the terms in parenthesees.

2.2x  =  -1.24x + 27.28 - 3.2

                                                     Add  1.24x  to both sides.

2.2x + 1.24x =  27.28 - 3.2

                                                     Combine like terms.

3.44x  =  24.08

                                                     Divide both sides by  3.44

x  =  7

Substitute  7  in for  x  in the second equation to find  y .

y  =  -0.4( 7 ) + 8.8

y  =  6

The maximum weight of a bowling ball doesn't matter for this question...maybe that is just a red herring.

We can find the surface area of a sphere only knowing its diameter.

surface area of sphere  =  4 π ( radius )²

radius  =  ( $\frac12$ )( diameter )  =  ( $\frac12$ )( 8 + $\frac12$ )  =  ( $\frac12$ )( $\frac{17}{2}$ )  =  $\frac{17}{4}$

surface area of sphere  =  4 π ( $\frac{17}{4}$ )²

surface area of sphere  =  4 π ( $\frac{289}{16}$ )

surface area of sphere  =  $\frac{289\pi}{4}$     (square inches)

x  is directly proportional to  y³     so     $x=ky^3$     where  k  is a constant.

y  is inversely proportional to  √z     so     $y=\dfrac{s}{\sqrt{z}}$     where  s  is a constant.

$x=ky^3\\~\\ x=k\Big(\frac{s}{\sqrt{z}}\Big)^3 \\~\\ x=\frac{ks^3}{z\sqrt{z}}$

When   z = 12 ,   x = 3     So...

$3=\frac{ks^3}{12\sqrt{12}}\\~\\ ks^3= 3\cdot12\sqrt{12}\\~\\ks^3=72\sqrt3$

When   z = 75 ,

$x=\frac{ks^3}{z\sqrt{z}}\\~\\ x=\frac{72\sqrt3}{75\sqrt{75}}\\~\\ x=\frac{72\sqrt3}{375\sqrt{3}}\\~\\ x=\frac{24}{125}$

CD  =  BD - BC

CD  =  22 cm - 6 cm

CD  =  16 cm

area of triangle ABC  =  ( 1/2 )( base )( height )

42 cm²  =  ( 1/2 )( 6 cm )( AC )

42 cm²  =  ( 3 cm )( AC )

14 cm  =  AC

Triangle  ACD  is a right triangle and now we know the length of both legs.

tan( angle )  =  opposite / adjacent

tan( x )  =  14 cm / 16 cm

tan( x )  =  7 / 8

x  =  arctan( 7/8 )

x  ≈  41.186°

9.)

$r\ =\ \frac{4}{2\cos\theta-3\sin\theta}\\~\\ r(2\cos\theta-3\sin\theta)\ =\ 4\\~\\ 2r\cos\theta-3r\sin\theta\ =\ 4\\~\\ 2x-3y\ =\ 4\qquad\ \qquad\ \qquad\text{because}\qquad x=r\cos\theta\qquad\text{and}\qquad y=r\sin\theta\\~\\ 2x\ =\ 4+3y\\~\\ 2x-4\ =\ 3y\\~\\ \frac23x-\frac43\ =\ y\\~\\ y\ =\ \frac23x-\frac43$

This is the equation of a line with a slope of   $\frac23$   and a y-intercept of  $-\frac43$ .

We are given a side-side-angle triangle, so we can use the Law of Sines to solve this.

$\frac{\sin B}{b}\ =\ \frac{\sin A}{a}\\~\\ \frac{\sin B}{4\sqrt{2}}\ =\ \frac{\sin 60°}{4\sqrt{3}}\\~\\ \sin B\ =\ \frac{\sin 60°}{4\sqrt{3}}\cdot4\sqrt2\\~\\ \sin B\ =\ \sin 60°\cdot\frac{4\sqrt2}{4\sqrt{3}}\\~\\ \sin B\ =\ \frac{\sqrt3}{2}\cdot\frac{4\sqrt2}{4\sqrt{3}}\\~\\ \sin B\ =\ \frac{\sqrt2}{2}\\~\\ \begin{array}{lcl} B\ =\ \arcsin(\frac{\sqrt2}{2})&\quad\text{or}\quad& B\ =\ 180°-\arcsin(\frac{\sqrt2}{2})\\~\\ B\ =\ 45°&\text{or}&B\ =\ 135°\\~\\ &&\text{But}\qquad135°+60°=195°>180°\\~\\ &&\text{So }\qquad B\neq135° \end{array}$

B  =  45°     is the only solution.

4.

y  is positive when  x  is in the interval  (-∞, -2)

y  is positive when  x  is in the interval  (0, 4)

3.

The yearly attendance at a ballpark is shown in the table.

Which answer describes the average rate of change from Year 2 to Year 5?

average rate of change from year  2  to year  5   =   $\frac{\text{333,700 people - 298,300 people}}{\text{5 years - 2 years}}$

average rate of change from year  2  to year  5   =   $\frac{\text{35,400 people}}{\text{3 years}}$

average rate of change from year  2  to year  5   =   $\frac{\text{11,800 people}}{\text{1 years}}$

average rate of change from year  2  to year  5   =   11,800 people per year