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1) The solutions are -4/3, -2/3.

2) The solutions are -1, -3.

HELP PLEASE ASAP! :)

3(2x + 4) = 2x - 4

6x + 12 = 2x - 4

4x = -16

x = -4

The equation form 1.2 of week 6 is 3xn+n-1

The point-slope is (Y-1)=3(x-1)

The slope-intercept is Y=3x-2

-2x   + 23  <  17    

-2x  <  17   - 23

-2x  <  - 6                divide both sides  by -2, reverse  the  inequality sign

x  >  3

y =  (3n) + ( n - 1) =  2n - 1

When  n  =  3.....then  y = 2(3)   -1   = 5

When n = 2  then  y  =  2(2) - 1   = 3

When n = 4 , then y = 2(4)  - 1

Using x  for n...the slope = 2

Using  ( 3 , 5)

Point-slope   is  (y - 5)  = 2(x - 3)

Slope - int   is    y = 2(x - 3) + 5  ⇒  y  = 2x - 6 + 5  ⇒ y=  2x  - 1 

Here's the graph and the points  : https://www.desmos.com/calculator/bgp0oajbn5

Let  those  only  in  A  (and not in the intersection)  = w

Let those only in B (and not in the intersection)  =  y

Let  those in the intersection =   x 

Let  those  not  in (A U  B)   =  z

So

 w +  x + y   =  31 =  A U B

 w  + x   =  17       

So

17  + y  = 31

y = 14

Those  not in A =   y  + z

y + z  =  36

14 + z = 36

z  =  22

Those not in B  = w + z

w  + z  = 33

w  + 22  = 33

w  =11

w  + x + y =  31

11  + x + 14  = 31

25 + x   = 31

x = 6

So

n(A)  =  11  + 6 =  17

n(B)  =  14 + 6   = 20

n(A ∩ B)  =  6

n(A U B)  = n(A)  + n(B)   - n (A ∩ B) =    17  + 20  - 6    = 31 

n(U)  =  n(A U B)  +  n(A U B)'  =    31  + 22   =   53

36 / 28 = q / 16 - q

36(16 - q) = 28q

576 - 36q = 28q

576 = 64q

q = 9

\(\frac{2x + 4}{5} = \frac{3x + 5}{7}\)

\(14x + 28 = 15x + 25\)

\(x = 3\)

AC - DC  = AD

20  - 5  =   15

AD  / BD =  BD / CD

15  / BD  = BD  / 5

15 * 5  = BD^2

75  = BD^2

AC^2 =  BD^2  + CD^2

AC^2  =  75  + 5^2

AC^2  =  75  + 25

AC^2  =100

AC  = 10

DE  / EF   =   XY /  YZ

10 /  8  =     ( x  - 1)   / 4

5/4  =   ( x - 1)  /  4            cross-multiply

4 * 5  = 4  ( x   -1)

20 =  4x  - 4

20 + 4 = 4x

24 = 4x

x   = 24/4  = 6

We don't really need  Heron's Formula  ( unless you're a glutton for punishment .....)

We  know  two sides and   an included  angle  between  these sides

Area =  (1/2)  product of  the sides  * sine of the included angle

So

Area  = (1/2) bc *sin A

Area  = (1/2)  (77 * 41)  sin 79°   =  1549  (sq units)

As  before....triangle  ABD is  right

AB^2  -  AD^2 =  BD^2

8^2   -  4^2  =   

64  -  16  =

48  =  BD^2

So

AD  / BD  = BD / DC

4 / BD  = BD / DC          cross-multiply

4 DC  =  BD^2

4DC =  48

DC = 48 / 4   =  12

AC =  AD   +  DC   =  4  + 12  =  16

Angle BAC  =  90   - 42  =  48°

Angle  LMN = 90   - 48   = 42°

Therefore.....   angle A =  angle L      and  angle   B = angle M

Therefore....  by  AA   triangle   ABC is similar to triangle LMN    (answer  A)

(2y - 1)   ( 4y^10  + 2y^9  + 4y^8  + 2y^7)  =

2y ( 4y^10  +2y^9  + 4y^8  + 2y^7)  -  1  ( 4y^10  + 2y^9  + 4y^8  + 2y^7)  =

8y^11  + 4y^10  + 8y^9  + 4y^8   -   4y^10  - 2y^9 - 4y^8  - 2y^7  =

8y^11 + 6y^9  - 2y^7

a=1;b=1;p=0; cycle:d=a*10+b;if(a*b%(a+b-1)==0, goto loop, goto next); loop:printd,", ",;p=p+1; next:b++;if(b<10, goto cycle, 0);b=1;a++;if(a<10, goto cycle, 0);print"Total = ",p

OUTPUT:

11 , 12 , 13 , 14 , 15 , 16 , 17 , 18 , 19 , 21 , 31 , 34 , 41 , 43 , 49 , 51 , 56 , 61 , 65 , 71 , 78 , 81 , 87 , 91 , 94 , Total =  25 such integers.

We  can  use  the Law of Cosines

a^2  = b^2  + c^2  - 2 ( b*c)  * cos 44°

a^2  = 33^2  + 19^2  - 2 ( 33 * 19) * cos 44°

a^2  =  1089  + 361 - 2 ( 627) *cos 44°

a^2 = 547.9

a = sqrt (547.9)   ≈   23.4   miles

Here is my attempt:
Let the distance to top of the hill =D

D/5 + (1/2)D/4 =T, solve for T
T = (13 D)/40 - Oliver's Time

D/4 + (1/2)D/5 =T, solve for T
T = (7 D)/20 - Lenny's Time

Set D = 1
Oliver's Time =13/40 =0.325 =3 1/13
Lenny's Time= 7/20  =0.35 =2 6/7

Conculsion:Lenny reaches the top of the hill first.
Proportion of their hiking time: 2 6/7:3 1/13 =1:1 + 1/13

since the two angles are on the same line, they are supplementary (they add up to 180 degrees).

so, following that, we can craft this equation:

9x + 17 = 180 - 37

9x + 17 = 143

9x = 143 - 17

9x = 126

x = 14

hope this helped! please let me know if you are confused about anything i did :)

This is a Supplementary Angle. This means that when you add the two angles it must equal 180 degrees. So.......

3x + 25 + 2x = 180

5x + 25 = 180

5x = 155

x = 31

since the two angles shown are on the same line, we can craft this equation:

4x - 140 = 180 - 80

4x - 140 = 100

4x = 100 + 140

4x = 240

x = 60

hope this helped! please let me know if you are confused about anything i did 

2, 4, and 5 are correct.

l-4l = 4 and l4l = 4

7x + 90 + 20 = 180

7x = 70

x = 10

5x + 2y = 90

5 * 10 + 2y = 90

2y = 90 - 50

y = 20

Triangle  BCD  is  right with  hypotenuse   BC  =12   and leg CD  =  3

So   BD^2  =  BC^2  - CD^2  =  12^2   - 3^2  =  144  -  9   =   135

And.....like the problem  you  submitted before

CD / BD  = BD / DA   cross-multiply

CD * DA  =  BD^2

3 * DA  = 135

DA  = 135 / 3 =  45

So   CA  = CD  + DA  =   3 + 45   =   48