hairyberry

$a+ar+a{r}^{2}+a{r}^{3} + \dots=4\\{a}^{3}+{a}^{3}{r}^{3}+{a}^{3}{r}^{6}+{a}^{3}{r}^{9}+\dots=10\\\text{Using the sum of infinite geometric series formula}:\\\begin{cases} \frac{a}{1-r}=4 \\ \frac{{a}^{3}}{1-{r}^{3}}=10 \end{cases} \\\text{Isolate a:}\\ \begin{cases} a=4(1-r) \\ {a}^{3}=10(1-{r}^{3}) \end{cases} \\\text{Substitute:}\\ 64{(1-r)}^{3}=10(1-{r}^{3})\\ 54{r}^{3}-192{r}^{2}+192r-54=0\\ (r-1)(9{r}^{2}-23r+9)=0\\ r=1, r=\frac{23+\sqrt{205}}{18}, r=\frac{23-\sqrt{205}}{18}\\ \text{take the range} -1 <1 \;\text{therefore}\; r=\frac{23-\sqrt{205}}{18}$

For two absolute values to sum to zero, we must have 

$|a|+|b|=0$, then $\begin{cases} a = 0 \\ b = 0 \end{cases}$, because $|x|\ge0$.

Following this logic, the only possible solution is $\begin{cases} x + y - 7 = 0 \\ 4x - y + 12 = 0 \end{cases}$.

 $5x+5=0$

$\begin{cases} x = -1\\y=8 \end{cases}$.

Therefore the ordered pair (-1, 8) is the only ordered pair that works.

$\begin{cases} {n \choose 1}*a = -6 \\ {n \choose 2}*a^2 = 12 \end{cases}$

$\begin{cases} na=-6 \\ \frac{n(n-1)}{2}*a*a=12 \end{cases}$

$\begin{cases} na=-6 \\ na*a(n-1)=24 \end{cases}$

$-6*a(n-1)=24$

$a(n-1)=-4$

$an-a=-4$

$\begin{cases} a = -2\\n=3 \end{cases}$

${(1-2x)}^{3}=1-6x+12{x}^{2}-8{x}^{3}$

$\bf{c=-8}$.

weight of sugar = s

weight of flour = f

$\begin{cases} 2s+3f \le 40 \\ f \le 6 + 2s \end{cases}$

$\begin{cases} (2s)+3f \le 40 \\ f - 6\le (2s) \end{cases}$

$f-6 + 3f \le 2s+3f\le40$

$4f-6\le 40$

$4f\le46$

$f\le\frac{23}{2}$

23/2 pounds is greatest solution, this occurs when $f-6=2s$, $s=\frac{11}{4}$. (always make sure to check the solution works with inequalities).

In response to your question:

A good way to do this is by the rational root theorem.

For any polynomial, $a_nx^n+a_{n-1}x^{n-1}\dots a_1x^1+a_0$, the rational roots can all be expressed as:$\pm\frac{\text{factor of }a_0}{\text{factor of }a_n}$.

Therefore in this polynomial, the only possible rational roots are $\pm 1, \pm11$. (factors of a₀ are 1, 11 and the only factor of a_n is 1)

Plug these values into the function, to see which ones produce zero. $\pm1$ is a really, really common ones, so always try those first.

Remainder theorem is always helpful. Remember if for a function $f(x)$, $f(a)=0$, then $x-a$ is a factor of $f(x)$.

Here is the workthrough of the first step:

Use rational root theorem, plug in 1 to $f(x)={x}^{3}+3x^{2}+7x-11$.

${1}^{3}+3*{1}^{2}+7*1-11=0$.

Therefore we know because $f(1) = 0$, then x-1 is a factor of $f(x)$, so we can factor it out.

The equation of the first circle with center (2,2) is:

${(x-2)}^{2}+{(y-2)}^{2}=4$

The equation of the second circle with center (17, 17) is:

${(x-17)}^{2}+{(y-17)}^{2}=289$

Here is a graph:

The closest distance between the two is the length of the line segment conncting the 2 centers of the cricles minus the two radii.

Therefore the closest distance is $\sqrt{(17-2)^2+(17-2)^2}-17-2=15\sqrt{2}-17-2$.

By Heron's formula, the area of the triangle is 

$\sqrt{23*(23-11)*(23-15)*(23-20)}=\sqrt{23*12*8*3}=\sqrt{6624}=12\sqrt{46}$.

Because side * altitude / 2 = area, then.

$\frac{20x}{2}=12\sqrt{46}$

$10x=12\sqrt{46}$

$x=\frac{6\sqrt{46}}{5}$.

$\because \triangle TSU \sim \triangle TPM$

$\therefore \angle TMP = \angle TUS = {89}^{\circ}$

$\because \angle STU = {29}^{\circ}, \angle TZU = {90}^{\circ}$

$\therefore \angle TUZ = 180-29-90={61}^{\circ}$

$\therefore \angle TMZ + \angle TUZ = {160}^{\circ}$.

All the children and all the adults sit together, so suppose on one side there sits 4 children, on the other sides sits 4 adults.

Suppose one of the children is Johnny.

We must always rotate the table so that Johnny is at the bottom of any orientation. 

Therefore, the children must always be the bottom of the table, in the orientation.

The children have 4! ways and the adults have 4! ways.

Therefore, there are 576 ways to be seated. 

1)

Choosing any 2 random people is 18 choose 2 = 153.

2)

Choosing Mark and 1 other person is just 17, because there are 17 other people. 

3)

Exclude mark essentially, so now we calculate 17 choose 2 = 136.

4)

136 (our answer to #3) + 17 (our answer to #2) = 153(our answer to #1). Why do you think that might be?