DS2011

Well the length of OA multiplied by 5 gives 30, because the scale factor is 5, so \(5x=30\), where x = length of OA

\(x = 6\), so OA is 6 units

Answer: 6 units

\(t^2+4t-96\)

t = -b/2a

t = -4/2

t = -2

Plugging in -2 for t, we get:

\(4-8-96\)

\(-100\)

The coordinates of the vertex are (-2,-100)

The smallest value of the quadratic is -100

Answer:  -100

\(x^2 + y^2 = 2x - 6y + 6 + 14x - 16y + 80\)

Graph:

We can find the circle radius and we see that it is:

 \(\sqrt{271}\)

Area: \(\pi r^2\)

Area = \(271\pi\), or \(851.371609123\)

Answer: 271π units squared or 851.37 units squared

\(3x^2 + 12x + 4 - x^3 + 8x - 5x + 7x^2 + 11 \)

We can simplify this:

\(-x^3+10x^2+15x+15\)

We have:

\(-x^3+10x^2+15x+15= a(x+b)^3+c(x+d)^2+ex+f\)

a must equal -1 and b must equal 0 to get -x^3

c must equal 10 and d must equal 0 to get 10x^2

e must equal 15

f must equal 15

Answer, -1 , 0 , 10 , 0 , 15 , 15

I think that this is the only set of numbers that can fill in the blanks, although it is possible that there could be other values for b and d, which could also change e or f, but probably this is the only answer.

First, we can put all the n's on one side:

\(2n^2+20n<64\)

We can plug in values of n now, and get

N          Answer

1               22

2               48

3               78

0               0

-1              -18

-2              -32

-3              -42

-4              -48

-5              -50

-6              -48

-7              -42

We can see a pattern, the numbers are going backwards.

-8              -32

-9               -18

-10             0

-11             22

-12             48

-13             78

The integers that satisfy the inequality are:

-12,-11,-10,-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2

Which is 15 numbers.

Answer: 15

Well, we knw that AE is twice as big as ED, and AE + ED = AD. 

We can wwrite an equation:

2x + x = 30, where 2x is AE and x is ED.

3x = 30

x = 10

AE = 2x, so:

AE = 20

Answer: 20

Well we know that x can not equal 0, because then we would have a number that is undefined, so the domain of the function f(x) is 

( - inf , 0 ) U ( 0 , inf )

Answer: ( - inf , 0 ) U ( 0 , inf )

\(2x^2+4x-1=x^2-8x+3\)

We can make this is the form  

\(ax^2+bx+c=0\)

We get:

\(x^2+12x-4=0\)

a = 1, b = 12, and c = -4.

Using the quadratic formula \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\), we get:

\(x = {-12 \pm \sqrt{12^2+16} \over 2}\)

\(x = {-12 \pm \sqrt{160} \over 2}\)

\(x = {-12 - \sqrt{160} \over 2}\), \(x = {-12 + \sqrt{160} \over 2}\)

Now we can square both values of x and then add them:

-151.894663844 + 0.10533615595 = -151.789327688

Answer: -151.789327688