CPhill

Let  x  be the number of minutes it would normally take to drive the route

Since

R * T  = D     and  D  is the same in both cases, we have that

30 ( x + 18)  = 45 (x  - 8)        simplify

30 x + 540  = 45 x - 360     rearrange as

540 + 360 = 45x - 30x

900 = 15x  divide both sides by 15

60  = x   

So....the distance  is  30 (60 + 18)  = 30 mph ( 78  min)   

But  we need to  convert 78 min to hours =  78/60  = 39/30 hr

So...the total distance is  30m/h ( 39/30 h)  = 39 miles 

HAHAHA!!!...I'm only a "genius" at sporadic periods.....!!!!!

Here's a pic :

Since two of the sides must be parallel, assume that AB  is parallel to CD

Drop a perpendicular from C to AB and let the intersection be the point E

Let EB  = x

Similarly, drop a perpendicular from D to AB  and let the intersection be the point F

Let FA  = 52 - (39 + x)  = 13 - x

So....the height  of the triangles  = height of the trapezoid

The  height of the triangle on the right  =√[12^2 - x^2] = CE

The height of the triangle on the left is √ [5^2 - (13 - x)^2 ] = DF

And, as stated,  CE  = DF

Equating the quantities under the roots, we have

12^2 - x^2  = 5^2  - (13 - x)^2

144 - x^2 = 25 - (169 - 26x + x^2)

144 - x^2 = 25 - 169 + 26x - x^2

144  = 25 - 169 +26x

144 = -144 + 26x

288 = 26x

x  = 288/26  = 144/13

So...the  height of the trapezoid = √[12^2 - (144/13)^2 ] = 60/13

And the area  =  (1/2)(height)(sum of the bases)  =

(1/2)(60/13) ( 39 + 52)  =

(1/2) (60/13) (91)  =

(1/2)(60) (91/13)  =

(1/2)(60)(7)  =

(1/2)(420)  =

210 units^2 

3x  - 2y  = 6    rearrange as

3x - 6  = 2y     divide through by 2

(3/2)x - 3  = y

A line perpendicular to this will have a slope of (-2/3)

If  this line has a y intercept  of 2, then the point (0,2)  is on the graph

So......the equation of the line is

y  = (-2/3)x + 2

To find the x intercept, let y  = 0

0  = (-2/3)x + 2

-2 = (-2/3)x

x  = -2(-3/2)  = 3

Here's a graph : https://www.desmos.com/calculator/fi2c8ubbdw

a.   If the side of the square is 7 times longer, then the area  is 7^2  = 49 times more

So.....49 * 1000   = 49000

a.  We have an octagon

There are 8 congruent triangles each  with an included angle of 45° between sides of  13.07 in

So...the total area  = 8 (1/2)(13.07)^2 (sin 45°)  ≈ 483 in^2

a. We have a trapezoid with  one base of 6    and a height of  8

The other base  = 6 + 2 + 8  (the triangle on the left has the same base as height )  =  16

The area  is  (1/2)height * (  sum of bases) =

(1/2)(8)(22)   =

88 cm^2

a. For the triangle on the right....the base is opposite a 30° angle so  its length is 1/2 * 6  = 3ft

The height of the triangle [ and the trapezoid]  =  3√3 ft

So...the area of the trapezoid  =

(1/2)(height) (sum of the bases)  =

(1/2)(3√3)(11 + 14)  =

(1/2)(3√3)(25)  =

(3/2)(25)√3 =

(75/2)√3 ft^2

Square  with a "radius"  of 11m

The "radius"  is the distance from the center of the square to any vertex

The area is  4(1/2)11^2  = 2 * 121  = 242 m^2

Pentagon with a side of 9m

The pentagon will be composed of 5 congruent triangles

The height of each is given  by  (9/2)/tan (36°)

So...the total area is

5 (1/2)(9)(9/2)/tan(36°)  =  (5)(9/2)^2/tan(36°)  ≈  139.4 m^2

Pentagon  with a radius of 5m

We have 5 congruent triangles with sides of 5 and an included angle of 72° between these sides

The area  =   5 (1/2) (5)^2sin (72°) ≈  59.4m^2

Hexagon with a side of 12 yd

We have six congruent equilateral triangles...each side  = 12 yd

The area is  6(1/2)(12)^2* √3/2  ≈  216√3  yds^2 ≈ 374.1 yds^2

Since this parabola turns upward....1 will still be in the range when, at a minimum,  (m , 1)  is the

vertex

We have the form   ax^2 + bx + c

To  find "m"  we   can use    -b/ [2a]  =       5 / (2 * 1)   =  5/2  = m

So  (5/2, 1)  is the vertex

So....subbing this into the function for x  and f(x)  to find " c,"  we have

1  = (5/2)^2  - 5(5/2) + c

1 = 25/4 - 25/2 + c

1  = 25/4 - 50/4 + c

1  = -25/4  + c

1 + 25/4  = c

29/4   =  c

Here's a graph to show this : https://www.desmos.com/calculator/gbhwg6nerl

 Check for yourself that if c > 29/4,  then  1 will be out of range

Thanks, Daisy.....!!!...here's another way to see this

Break-even point is where  Costs  = Revenue

Let the number of screwdrivers that need to be sold  = S

So...Total  Costs = 124  +  1S

Total Revenue  = 3S

Set these even

124 + 1S  = 3S     subtract 1S from both sides

124  = 2S    divide both sides by 2

62  =  S   ⇒   62 screwdrivers must be sold to break even

I get the same answer as Melody....but....I  don't know if the included pic is accurate or not

I have assumed that  angle ABC  =  90 =  angle BAD

Since AC  bisects  angle BAD, then angle BAC =  angle CAD  = 45

AB = BC

So... AC  = √2AB  = CD

Area   of ABCD   =  Area  of triangle ABC   + Area of triangle ACD  = 42

So

Area  of triangle ABC  + area of triangle ACD  = 42

(1/2)AB*BC   + (1/2)AC * CD  = 42

(1/2)AB^2  + (1/2)√2AB * √2AB   = 42

AB^2  + 2AB^2  = 84

3AB^2  = 84

AB^2 = 28  = area of triangle ACD