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for this expression to be undefined, y^2-3y-2=0

sum of equation of the form ax^2+bx+c is -b/a (product is c/a)

sum of this equation is 3

JP

1/4x+25=0

multiply by 4 gives x+100=0

subtract both sides by 100 gives x=100

(b) There are 108/3 = 36 unique sets.

(c) For the first card, there are 3 ways to choose the number.  For the second, there 2 choices, and then there is only 1 choice.  So there are 3*2*1 = 6 ways that the numbers can be chosen.  Doing this for the other attributes, we get 6*6*24*6 ways.  But the order of cards doesn't matter in a set, so we divide by 3!: 6*6*24*6/3! = 864.  So there are 864 sets for part (c).

(d) The cards can have the same number, color, shape, or shading.  If all the colors are the same, then there are 6*24*6/3! = 144 sets.  If all the numbers are the same, then there are 144 sets.  We get the same number for shape and shading, so there are 4*144 = 576 sets for part (d).

(e) We need to choose two of the attributes.  There are C(4,2) = 6 ways of choosing two attributes.  For each of these two attributes, there are 3 options.  For the other two attributes, there are 3 ways of assigning the choices, so there are 6*3*3*3*3 = 486 sets for part (e)

(f) First we choose which attributes are the same.  There are C(4,3) = 4 ways of choosing which attributes are the same.  There are then 3*3*4 = 36 ways to assign which is which for each of these three attributes, and there are 4 ways to assign the choices for the fourth attribute, so there are 4*36*4 = 576 sets for part (f).

\(P(G(a)) = 4+ 2 \sqrt {4+8a+2} \)          defined

\(4+8a+2\ge 0 \)

\(6\ge -8a\)

\(8a\ge -6\)

\(a \ge (-6 /8) = -(3/4) \)

P(G(a)) is defined  iff  \(a \ge (-3/4)\)

So  a   has no upper bound ,  only lower bound.

There are 26 integers x that work.

The value of m is 101.
In proof, first note that $100^0 \equiv 1 \pmod{101}$.
Also, for $i\ge0$, we have $100^{i+1} = 100\cdot 100^i \equiv (101-1)(100^i) \equiv -(100^i) \pmod{101}$.

Let $n=d_kd_{k-1}\ldots d_1 d_0$ be a positive integer written in base-10,
where $d_i$ is a number (written in base-10) and $0\le d_i \le 99$
for all $0\le i\le k$.
For example, if $n=354764$ then $k=2$, $d_2 = 35$, $d_1=47$, and
$d_0=64$. We have
$$
n = \sum_{0\le i\le k} d_i \equiv
\sum_{\substack{0\le i\le k\\ i\rm{\ even}}} d_i +
\sum_{\substack{0\le i\le k\\ i\rm{\ odd}}} d_i
\equiv
\sum_{\substack{0\le i\le k\\ i\rm{\ even}}} d_i -
\sum_{\substack{0\le i\le k\\ i\rm{\ odd}}} d_i
\pmod{101}.
$$

It really just comes with time and practice. The denominator is very easy, assume we have ax^2+bx+c. Right away the denimonator is 2*a, that's it! The numorater is harder, think of it as:

Minus the middle coeficient (-b) 

Plus Minus 

Sqrt of ((b again :D)^2   -   (4*(The other two coefficents not mentioned before in the numerator (a*c)))

Hope that helps!

We have
$x^{10}-8x^8+7x^4+8x^3-12x^2-15x-5 = (x^2-1)(x^8-7x^6-7x^4+8x-12)-7x-17$.
So the remainder is $-7x-17$.

This sequence is called the Fibonacci sequence.
Let's write down and study the sequence of remainders
of the Fibonacci numbers modulo 4.
We have $1, 1, 2, 3, 1, 0, 1, 1, \dots$ so this sequence
of remainders is periodic where every 6th term only is divisible by 4.
Therefore, of the first 1000 Fibonacci numbers, $\lfloor 1000/6 \rfloor = 166$ of them are divisible by 4.

You're back!

Here is a picture....   You are trying to find the green line distance   ( Hint: Cosine law)

Thank you, why do you add an x to each of their ages?  I thought x was the number of years that we are looking for.

Thank you, why do you add an x to each of their ages?  I thought x was the number of years that we are looking for.

There's a way to simply brute force this, so it's not pretty, but it should be the right answer.

If each set of integers are three digits, and subtracting X from Y gives 333, we can assume that subtracting the single-digits in X (_1/2/3) from Y (_1/2/3) should yield 3. There's only three ways to get 3 using the numbers 1 through 6 with solely subtraction: 6-3, 5-2, and 4-1. We can therefore assume X (_1/2/3) = 4, 5, and 6; and Y (_1/2/3) = 1, 2, and 3.

It's a simple matter of lining up the ones, tens, and hundreds columns so that each column will spit out a 3. You have three columns, and three problem sets to apply to each column, which should result in nine different values for X/Y to go together to fulfill X - Y = 333.

I won't solve the whole thing for you, as I'm not sure whether the problem asks for fully simplified answers or just the problem sets, but I'll at least give you that little stepping stone.

$-3(1+4i) + i(-2-i) = -3 -12i -2i - i^2$.

We know that $i^2 = -1$, therefore, this equation is now $-14i - 3 - (-1) = \boxed{-2 -14i}$, which is our final equation. 

It is not defined when the denominator is 0. 

x^2 - 24x + 100 = 0

Solve for x. :))

=^._.^=

Let $x=\sqrt{1+\sqrt6}$.  Then $x^4=7+2\sqrt6$, that is, $x^4-7=2\sqrt6$,
so that $(x^4-7)^2 =24$, so that $(x^4-7)^2-24=0$.
This means $\sqrt{1+\sqrt6}$ is a zero of the polynomial
$q(x)=(x^4-7)^2-24$.
Now $q(x)=(x^4-5-2x^2)(x^4-5+2x^2)$.   Let $p(x) = x^4-5-2x^2$.
We then find that $\sqrt{1+\sqrt6}$ is a zero of $p(x)$, and $p(x)$ is a
degree-4 polynomial with integer coefficients and leading coefficient 1.
It's given that such a polynomial $p$ is unique. Thus,
$p(1)=1^4-5-2\cdot 1^2= 1-5-2=-6$.

Thank you, I apologize for assuming you were wrong. I'm logged out.

Either

2 ( 2d - 40)  +  (d + 10)  =   180

4d  -  80  + d + 10   =  180

5d  -  70  =   180

5d  =  180 + 70

5d  =  250

d  =  250 /  5    =   50

or  possibly

2 ( d + 10)  +  2d - 40   =   180

2d + 20  + 2d  - 40   =  180

4d  - 20  =  180

4d   = 180 + 20

4d  = 200

d  =   200  / 4    =  50 

d  = 50

Answers  in red

Let x  be  the  number of years we are looking for

(4 + x)  + ( 2 + x)   =  (30 + x)  /  2

6 + 2x  =   (30 + x)  / 2           multiply through by 2

12 + 4x  =  30 + x

4x - x  =   30  -  12

3x  =  18

x  =  18 / 3    =   6 yrs

Proof

10   + 8  = 18

36/2  =  18