proyaop

\(4 = a + ar + ar^2 + ar^3 + \dotsb\)

\(10 = a^3 + a^3r^3 + a^3r^6 + \dotsb\)  Find \(r\)

Infinite geometric series formula: a1 [first term] divided by (1 - r) [1 - common ratio] = \(a_1\over{(1 - r)}\)

Using this piece of information, we can rewrite the equations from earlier:

\(4 = {a\over{(1 - r)}}\); \(4 - 4r = a\)

\(10 = {a^3\over{1 - r^3}}\); \(10 - 10r^3=a^3=(4-4r)^3=-64r^3+192r^2-192r+64\)

\(10 - 10r^3 = -64r^3+192r^2-192r+64\); \(54r^3-192r^2+192r-54=0\); \(54r^3-54 - 192r^2+192r=0\)

Factor by grouping: This part is hard...

\(54(r^3-1)-192r(r-1)=0\)

\(r^3-1=(r-1)(r^2+r+1)\); remember this!!! Factoring difference/sum of cubes is important!!

\(54(r-1)(r^2+r+1)-192(r-1)=0\); \((r - 1)(54r^2+54r-138)=0\)

We know the common ratio cannot be 1, because then the series would not converge.

\(54r^2+54r-138=0\); divide by 54: \(r^2+r-{23\over9}=0\)

By quadratic formula:

\(r = {-1 \pm \sqrt{1+{92\over9}} \over 2}={-1\pm{\sqrt{101}\over3}\over2}=-{1\over2}\pm{\sqrt{101}\over6}\)

please reply if you catch any errors... because this number is ugly :/

Rather than expanding the entire thing, just expand the terms that would multiply to a term with x^2:

(x^3 + x^2 + x + 1)(x^4 + 3x^3 + 8x + 7x + 1) - if you had a typo here where 8x is supposed to be 8x^2, don't worry and scroll down.

Expanding out the only terms that would give x^2

x^2 * (+1) + x * (8x + 7x) + 1 * (nothing) = x^2 + 15x^2 = coefficient of 16

Solution for if there was a typo:

x^2 * (+1) + x * (7x) + 1 * (8x^2) = x^2 + 7x^2 + 8x^2 = 16x^2 = coefficient of 16 (it's the same dw)...

You can generally do the box method or the X method, with the multiples and the coefficients multiplying each other and then summing to get the B term: However, there are other couple notable methods for factoring quadratics:

For example, if you notice there is no B term, you can try difference of squares if c is negative

Another example is factor by grouping. You can do this by splitting up the B term into two different terms, and then take out the GCF of each group, and then using distributive property (but reversed).

For example: 2x^2 + 5x + 2.

2x^2 + 4x + x + 2

2x(x + 2) + 1(x + 2)

(2x + 1)(x + 2)

I hope you don't mean factoring as vertex form or expanded form... 

ax^2 + bx + c = 0

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

Complementary = two angles sum to 90 degrees

Supplementary = two angles sum to 180 degrees

So, angle F + angle G = 90 degrees; angle F is 26 degrees:

26 degrees + angle G = 90 degrees

angle G = 90 degrees - 26 degrees

angle G = 64 degrees

\(\frac{1}{1 \times 4} + \frac{1}{4 \times 7} + \frac{1}{7 \times 10}={1\over{4}}+{1\over28}+{1\over{70}}\)

To add fractions, multiply the denominator by a certain number so that the three denominators have a "common multiple".

From the factors, we can see that the Least Common Multiple of 4 = 2^2, 28 = 2^2 * 7, 70 = 2*5*7, is 2^2 * 5 * 7 = 140 (taking the highest exponents--- all this is basic and stuff that you should know already).

So: \(={35\over140}+{5\over{140}}+{2\over140}={35+5+2\over{140}}={42\over140}={3\over{10}}\)

\(\frac{3-z}{z+1} \ge 2(z + 4)\)

Case 1: z + 1 is greater than 0, so when you multiply both sides by z + 1, the inequality sign will NOT flip. z + 1 > 0, z > -1. (z cannot be -1, or else denominator will be 0 and thus an undefined expression on the left side of the inequality...)

\(3-z{\ge}2(z+4)(z+1)\)

\(3-z{\ge}2z^2+10z+8\)

\(2z^2+11z+5{\le}0\)

\((2z+1)(z+5){\le}0\)

z has to be [-5, -1/2] to satisfy the inequality. However, we know z > -1, so the domain is (-1, -1/2].

*(2z + 1)(z + 5) have to be one positive * one negative (or one is 0) to be less than equal to 0, so by picking [-5, -1/2], we can ensure their sign + or - is always the opposite (except at 0...)*

Case 2: z + 1 is less than 0, so when you multiply both sides by z + 1, the inequality sign WILL flip. z + 1 < 0, z < -1.

Follow the factoring from above...

\((2z+1)(z+5){\ge}0\)

Both factors have to be either both positive or both negative, so z has to be (-inf, -5] U [-1/2, inf) to satisfy the inequality. However, we know that z < -1, so we take the left side of the union; (-inf, -5].

Combining the two cases (joining their unions) would be (-inf, -5] U (-1, -1/2].

Combine like terms:

\(4t^2{\le}-13t+35+4t^2-1\)

\(13t{\le}34\)

\(t{\le}{34\over{13}}\)

Your answer in interval notation would be (-inf, 34/13]

\((x-2)(x+6)\le(x-2)(x+5)\)

\(x^2+4x-12 {{\le}} x^2+3x-10\)

\(4x-12{\le}3x-10\)

\(x{\le}2\)

Be careful when you divide by x - 2, because it can remove a possible solution, and also mess with the inequality sign!

Your answer in interval notation is (-inf, 2]