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p(x) + (x⁵ + 3x³ + 9x)  =  (7x³ + 24x² + 25x + 1)

Subtract  (x⁵ + 3x³ + 9x)  from both sides of the equation.

p(x)  =  (7x³ + 24x² + 25x + 1)  -  (x⁵ + 3x³ + 9x)

p(x)  =  7x³ + 24x² + 25x + 1 - x⁵ - 3x³ - 9x

p(x)  =  -x⁵ + 7x³ - 3x³ + 24x² + 25x - 9x + 1

p(x)  =  -x⁵ + 4x³ + 24x² + 16x + 1

Parallel lines have the same slope.

If the lines are parallel, then the slopes of the lines are equal.

If the lines area parallel, then....

\(\frac3n\,=\,-\frac72 \\~\\ 3\cdot2\,=\,-7\cdot n \\~\\ 6\,=\,-7n\\~\\ n\,=\,-\frac67\)

....You're right!!!

If the lines are perpendicular, then the slopes are negative reciprocals of each other.

That means   \(\frac3n\)   must equal the negative reciprocal of   \(-\frac72\)  .

The negative reciprocal of   \(-\frac72\)   is    \(\frac27\)  .

If the lines are perpendicular, then....

\(\frac3n\,=\,\frac27\\~\\ 7\cdot3\,=\,2\cdot n\\~\\ 21\,=\,2n\\~\\ n\,=\,\frac{21}{2}\)

area of stone path in sq ft   =   16x  +  8x  -  x²

area of stone path in sq ft   =   24x - x²        Plug in  80  for the area of the stone path.

80  =   24x - x²        Add  x²  to both sides of the equation

x² + 80   =   24x      Subtract  24x  from both sides.

x² - 24x + 80   =   0       What two numbers add to  -24  and multiply to  80  ?  -20  and -4 .

(x - 20)(x - 4)   =   0       Set each factor equal to  0  and solve for  x .

x - 20  =  0         or        x - 4  =  0

x  =  20              or        x  =  4

The width of the paths must be less than the lengths of the sides of the big rectangle,

so  x  must be less than  16 , and  x  must be less than  8 .

Since  20  is too large for the width of the paths, the only solution is  x  = 4

The "reciprocal" of a number is just 1 divided by that number.

The reciprocal of   \(x\)   is   \(\frac1x\) .     The reciprocal of   \(\frac{a}{b}\)   is   \(\frac{b}{a}\) .

But the slopes of perpendicular lines are not just reciprocals of each other.

The slopes of perpendicular lines are negative reciprocals of each other. To find the slope of a line perpendicular to one with a given slope, we must take the negative reciprocal of the given slope. That means take the reciprocal of the slope and also multiply it by -1 .

For example...

If a line has a slope of  \(\frac34\) ,  the slope of a line perpendicuar  =  -\(\frac43\)

For a shape to be inscribed in a circle, all the vertices have to be on the circle.

A square can be inscribed in a circle, like this:

Since a square is a rectangle, and a square can be inscribed in a circle, a rectangle can be inscribed in a circle. And since a square is a parallelogram, and a square can be inscribed in a circle, a parallelogram can be inscribed in a circle.

Note that a rectangle with different side lengths could also be inscribed in a circle. However, a parallelogram with non-right angles could not be inscribed in a circle.

An isosceles trapezoid can also be inscribed in a circle, like this:

All of the options could be inscribed in a circle.

\(x^a\cdot x^b\,=\,x^{a+b}\)

So....

\(\sqrt3^\sqrt2\cdot\sqrt3^{-\sqrt2}\,=\,\sqrt3^{\sqrt2+-\sqrt2}\\~\\ \phantom{\sqrt3^\sqrt2\cdot\sqrt3^{-\sqrt2}}\,=\,\sqrt3^{\sqrt2-\sqrt2}\\~\\ \phantom{\sqrt3^\sqrt2\cdot\sqrt3^{-\sqrt2}}\,=\,\sqrt3^0\\~\\ \phantom{\sqrt3^\sqrt2\cdot\sqrt3^{-\sqrt2}}\,=\,1\)

I THINK the graph of  y = f(x)  looks something like this.....

f’(0)  =  0  so  when  x = 0  the slope of f(x) is  0

f’(1)  =  3  so  when  x = 1  the slope of f(x) is  3

f’(3)  =  1  so  when  x = 3  the slope of f(x) is  1

When  x  is between 0  and  1  the slope is smaller, closer to zero, than it is when  x = 1 .

Notice that there is a value between  0  and  1  that also makes the slope be  1 .

When  x  is greater than  1  the slope is closer to zero than it is when  x = 1 .

f’(-1)  =  -3  so  when  x = -1  the slope of f(x) is  -3

f’(-3)  =  -1  so  when  x = -3  the slope of f(x) is  -1

When  x  is between  0  and  -1  the slope is closer to zero than it is when  x = -1 .

When  x  is less than  -1  the slope is closer to zero than it is when  x = -1 .

This picture may help:

Remember that we want to find the area of the walk, that is the shaded area in the picture.

You found the area of a circle with a radius of 8.3 .

Below is the full answer......

area of walk   =   area of outer circle - area of inner circle

radius of inner circle in feet   =   12.4 / 2   =   6.2

area of inner circle in sq feet  =   pi * 6.2²   =   38.44pi

radius of outer circle in feet   =   6.2 + 2.1   =   8.3

area of outer circle in sq feet   =   pi * 8.3²   =   68.89pi

area of walk in sq feet   =   68.89pi - 38.44pi   =   30.45pi   ≈   95.7

Let the length of the dashed line be  x .

The sum of the lengths of any two sides in a triangle must be greater than the length of the third side.

From the top triangle, we can say...

8 + 10  >  x

18  >  x

x  <  18

8 + x  >  10

x  >  2

10 + x  >  8

x  >  -2

From the bottom triangle, we can say....

12 + 16  >  x

28  >  x

x  <  28

12 + x  >  16

x  >  4

16 + x  >  12

x  >  -4

So  x  has the restrictions:    x  <  18 ,   x  >  2 ,   x  >  -2 ,   x  <  28 ,   x  >  4 ,   x  >  -4

That is....    x  <  18   and  x  >  4

So the whole numbers that  x  can be are:

5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17

There are  13  different whole numbers that could be the length of the dashed line.