doeansodc

the sequence goes as following:

2, 7, 12, 17, ...

the sequence of sums goes:

2, 9, 11, 28, 50, 77, 105

The seventh term sends Shirley running from the room.

That number is 2+6*5=32

For a system of equations to have infinitely many solutions, the equations should be the same. We start by simplifying:

$8a+b=-3$

$9a+(2-m)b=k$

now we make the a terms the same:

$8a+\frac{8}{9}*(2-m)b=\frac{8k}{9}$

this means

 $\frac{8k}{9}=-3, so\: k=-\frac{27}{8}$

and

$2-m=1, so\:m=1$

The easiest way to solve this problem is with complementary counting.

There are 2 ways for the coin to be tossed less than or equal to 4 times:

1. 3 in a row for heads

There is a (1/2)^3 probability for this to happen in 3 tosses and a (1/2)^4 chance for it to happen in 4.

2. 3 in a row for tails

This is the same as case 1, just with heads and tails reversed.

Adding up all the ways, it is 2(1/8+1/16) or 3/8 chance for the coin to be tossed less than or equal to 4 times.

Subtracting it from 1 gives 5/8, the odds for the coin to be tossed more than 4 times.

expand:

x^2+6x>16-x+14+x^2

cancel out x^2:

7x>30

x>30/7

in interval notation:

(30/7, ∞)

edited i made some silly mistakes

we can start by moving all the terms of the equation onto the left side: 

x^2-14x+55-6y+y^2=0

then we can factor:

(x^2-14x+49)+(y^2-6y+9)-3=0

(x-7)^2+(y-3)^2=3

this gives us the equation of a circle with center (7, 3) and radius √3

the circle is always below the line, so we can just find the area of the circle, which is just πr^2, which gives us 3π