CPhill

The keys to  solving  these

(1)  line up all variables  on  one side of the equation and constants on the other

(2)  pick a variable to eliminate....either will do....this is the REAL key

(3) not a necessity...but I always like to get rid of fractions, too, before  I start

7x + 2y  = -10

-14x  - 3y  = 29

I see that if I multiply the top equation by  2 and add it to the second equation,  the "x" term will disappear..so

14x  + 4 y  = -20

-14x - 3y  = 29    add these  and we get

y  = 9

To find x...we can  use any equation...so...

7x + 2(9)  = -10

7x + 18  = -10

7x = -28

x  = -4

So.....(x, y)  =(-4, 9)

Do you want to try a couple on your  own, RainbowPanda  ??

A vertical asymptote will  potentially  occur  where an x value makes the denominator of this rational function  = 0

To find these possibilities....set the denominator   =  0

x^2 + 6x  - 7   =  0

(x + 7) ( x - 1)    = 0

Setting both factors to 0 and solving for x produces   x  = -7  and  x  = 1

However.... since the  numerator  also  has the  factor (x -1), we will have a "hole"  at x  = 1

The only  true vertcal asymptote    will be at   x  =  -7

See the graph  here :  https://www.desmos.com/calculator/ohkiuoni6o

Tough to tell, NSS...but it looks like 1997 - 1998 

We are really  just looking for the countries where these two conditions exist :  the "orange" bar  is  "longer" than the " blue" bar  and the "gray" bar  is "longer" than the "orange" bar

Three  qualify  -  Australia, Canada and Egypt 

1. The first and thirteenth terms of an arithmetic sequence are  7/9  and 4/5 , respectively. What is the seventh term?

We have that

4/5  =  7/9 + d (12)

4/5 - 7/9 =  d (12)

1/45  = d(12)

1 (45 *12)  = d

1/540  = d

So...the seventh term is....    7/9  + (1/540)(6)    = 71 / 90

The common ratio, r, can be found  by   [ successive term  /  preceding term ]  = 16 / 12 =  4/3

Since...to get the third term...we had to multiply the first term  by  "r"  twice...se have this equation ;

(first term)  * r^2  = 12

(first term) (4/3)^2  = 12

(first term) ( 16/9)   = 12     multiply both sides by   9/16

first term   =  12 (9/16)   = (12/16) (9) = (3/4)(9) =  27/4

After the multiples of 2 are removed....there are 25  integers left

We are looking for the odd multiples of 3  [ the even ones have been accounted for ]  =   3 , 9 , 15, 21, 27, 33, 39, 45

Ther are 8  of these...so

25  - 8   = 17  integers   in set  S

We have two  equations in two unknowns...call the common  difference, d

(1/2) +  d  = x -1  ⇒    (1/2) + 1  = x - d      (1)

x -1 + d  = 3x    ⇒       -1 =    2x  - d    ⇒  1 = -2x + d      (2)

Add (1)  and (2)  and we get

(5/2)  = - x

-5/2  =  x

And  d  =  (3/2)  = (-5/2) - d ⇒   d  =  -5/2  - 3/2  = -8/2   = -4

Proof

1/2 , -5/2  -1 , 3 (-5/2)

1/2 , -7/2 , -15/2         with   d  =  -4

Since one angle is  50°, then   x  = [ 180 - 50  ]  / 2  =   130  / 2   = 65°

Since two angles  = x, then these angles are equal

The triangle is

(a) acute  -  all angle are  < 90°

(b) isosceles -  two angles are equal, therefore the two sides opposite them are equal

Second one

y <  (-3/4) x + 1

This is a  dashed line - because of the "<"  - with a negative slope  and a y  intercept of  1

y > (2/3)x - 3

This  is also a dashed line with a positive slope and a y intercept of -3

Note that the graph   on the botom right is the only one satisfying these conditions....so, by default...it must be the  correct one