WhyamIdoingthis

All of the values of x from [-25, -sqrt 5) would get squared and turned positive, and outside of the domain. So they don't work.

All of the values of x from (sqrt 5, 5] would get squared outside of the domain of f(x). So they don't work.

Everything inside those still works, so the width of the interval is $2\sqrt5$.

ab has to be a perfect square.

b has to be a perfect square.

(4 + sqrt(4b) + b)/3 = (4 + 2sqrt(b) + b)/3 = an integer

And from there you just plug values in.

9 doesn't work.

16 doesn't work.

25 works.

So the value of b that answers this problem is 25.

You can combine these two, since the common ratio is the same.

6, a, b, 1296.

1296 = 6 x r^3

216 = r^3

r = 6

a = 6x6 = 36

This needs to be minimized: ${13\over 1-r}$.

Where -1 < r < 1. (If that isn't true then the series doesn't really have a sum)

If r is -1, the sum is 13/2 = 6.5. Rounding this figure up will both give an integer and put r back into its accepted range, so the smallest integer that can possibly be the sum of an infinite geometric series whose first term is 13 is 7. (the common ratio would be -6/7)

${d+1\over d} \times {d+2\over d+1} ... \times {d+20\over d+19}=3$

The numerator of the first term cancels the denominator of the second, and so on until the numerator of the 19th term cancels the denominator of the 20th.

You are left with d+20/d =3

3d = d + 20

2d = 20

d = 10

The original fraction is d+1/d = 11/10

If the parabola opens upwards, then a must be positive.

The roots will always be 3 and -2. This gives a base of 5.

5h/2 = 12

h = 24/5

However, since the parabola opens upwards, the intersection with the y-axis will be negative.

The graph of the parabola must have a y intersect at -24/5. Plugging this point in:

-24/5=a(-3)(2)

-24/5=-6a

4/5 = a

This is true: $a_n^2 - a_{n + 1} = (-1)^n$

And we have $a_1 = 5$

So this must be true: $5^2-a_2=(-1)^1$, or $26 = a_2$.

Similar logic:

$26^2-a_3=(-1)^2$

$\fbox{$a_3=675$}$

x * (100-20)% = 500

x * 80% = 500

4/5 * x = 500

4x = 2500

x = 625

(All of these values are found using simple area formulas, or length formulas derived from the Pythagorean Theorem)

AB is the square root of 2.

The area of the semicircle is pi/2.

The area of the square is 2.

The area of the circle around the square is pi.

The area of the part of the circle between arc AB and AB is (pi-2)/4

The area of the lune (shaded part) is pi/2 - (pi-2)/4 = pi/4 + 1/2

(I think)

The sidelength of the square inscribed in ABC has to be half of BC.

The sidelength of the square inscribed in DEF has to be a third of DF.

The ratio of BC:DF is 1:square root of 2

The ratio of 1/2:1/3 is 3:2

The ratio of the sidelength of the square inscribed in ABC to that of the square inscribed in DEF is 3:2 times the square root of 2.

Therefore, the ratio of their areas is 9:8.

 15 x 8/9 = 40/3

f(2) = -2. 

I don't think this requires explanation.

The vertex will come three units to the left compared to the original.

(-1, -1)

x^2-6x+k=0

$x = {-(-6) \pm \sqrt{(-6)^2-4(1)(k)} \over 2(1)}=3\pm1\sqrt{9-k}$

If k gets any bigger than 9, there both roots will be zero. Hence the largest value k can be such that that equation has at least one real solution is 9.

Q(x)=(lx^2+mx+n)

Q(x)^2=(l^2x^2 + 2mlx^3 + (2ln+m^2)x^2+2mnx+n^2)

Therefore

l^2=1

2ml=2

ml=1

2ln+m^2=-4

a=2mn

b=n^2

If l is one, so is m.

If l is -1, so is m.

l has to be negative, because of 2ln+m^2=-4, which means 2ln is negative.

2ln=-5

n=2.5

b=6.25

a=-5

a+b=1.25

(I think)

a(8-3x)+b=x

-3ax+8a+b=x

a would need to be -1/3 in order to keep the x term on the left isolated. 8a+b would need to be equal to 0, so be would be 8/3. 8/3+(-1/3)=7/3.

I assume $N$ is for natural numbers and $Z$ is for integers.

1.

A = {1, 2, 3, 4, 5, 6}

B = {4, 7, 10, 13}

Subsets of B are any combination of the elements of B (excluding some).

2. 

If A is contained within B, and B is contained with C, A is contained within C. 

This is, I suppose, one of those annoying proofs which I take for granted and see no way of prooving it other than it is logical. It is here I should also note that I have never studied set theory, and I'm probably getting everything wrong .

3.

{4(3a+b)|a,b $\in$Z} = {4c|c$\in$Z}

{3a+b|a,b$\in$Z}={c|c$\in$Z}

3a+b$\in$Z

c$\in$Z

(I don't know what to do after that....)

Again, I've never studied this, so apologies in advance for faulty logic everywhere.

Answer: $4\sqrt{1486}\over{-49}$

Solution:

h(t) should be equal to 4.

So:

-4.9t² + 18t - 0.4 = 4

-4.9t² + 18t - 4.4 = 0

Putting this into the quadratic formula gives:

$-18\pm\sqrt{237.76}\over{-9.8}$

You need the bigger one of these roots minus the smaller one, which is equal to 2 times the square root of 237.76 over -9.8. This is also equal to $4\sqrt{1486}\over{-49}$.

$T(b+1) = \frac{(b+1)(b+2)}{2}=\frac{b^2+3b+2}{2}$

$T(b) = \frac{b(b+1)}{2}=\frac{b^2+b}{2}$

$T(b+1)-T(b)=b+1$

$b+1$ should be a triangular number.

The smallest triangular number after 2011 is 2016 (take 2011, double it, square root the result, round that, plug it in as n. It will at the very least get you close to the triangular number wanted and then you will have know that number is the nth triangular number, making finding the next easy).

$b+1=2016$

$b=2015$