heureka

The taylor family and the simmons family each used sprinklers last summer.
The water output rate for the taylor family's sprinkler was \(40~l\) per hour.
The water output rate for the simmons family's sprinkler was \(35~l\) per hour.
The families used their sprinklers for a combined total of \(35~ hours\), resulting in a water output of \(1300~l\).
How long was each sprinkler used?

I assume:

\(\begin{array}{|rcll|} \hline \dfrac{40~l}{h}*t_1 + \dfrac{35~l}{h}*t_2 &=& 1300~ l \\ && t_1+t_2 = 35~ h \\ && \text{or} \quad \boxed{t_2 = 35~h-t_1} \\\\ \dfrac{40~l}{h}*t_1 + \dfrac{35~l}{h}*(35~h-t_1) &=& 1300~ l \\\\ \dfrac{40~l}{h}*t_1 + 35^2~l-\dfrac{35~l}{h}*t_1 &=& 1300~ l \\\\ \dfrac{40~l}{h}*t_1-\dfrac{35~l}{h}*t_1 &=& 1300~l-35^2~l \\\\ \dfrac{5~l}{h}*t_1 &=& 75~l \\\\ t_1 &=& \dfrac{75~l}{5~l}~h \\\\ \mathbf{t_1} &=& \mathbf{15~h} \\ \hline t_2 &=& 35~h-t_1 \\ t_2 &=& 35~h-15~h \\ \mathbf{t_2} &=& \mathbf{20~h} \\ \hline \end{array}\)

taylor family's sprinkler was \(\mathbf{15~h}\) used.
simmons family's sprinkler was \(\mathbf{20~h}\) used.

For what real values of $c$ is $9x^2 + 16x + c$ the square of a binomial?
If you find more than one, then list your values in increasing order, separated by commas.

\(\begin{array}{|rcll|} \hline \left(3x+\dfrac{8}{3}\right)^2 &=& 9x^2 + 16x + \dfrac{64}{9} \quad & | \quad c=\dfrac{64}{9} \\ \hline \end{array}\)

For what real values of $a$ is $x^2 + ax + 25$ the square of a binomial?
If you find more than one, then list your values in increasing order, separated by commas.

\(\begin{array}{|rcll|} \hline (x+5)^2 &=& x^2 + 10x + 25 \quad & | \quad a=10 \\ (x-5)^2 &=& x^2 - 10x + 25 \quad & | \quad a=-10 \\ \hline \end{array} \)

According to Pythagoras theorem, distance between points (-3, 8) and (8, -5) is

\(\text{Let $P_1(x_1,~y_1)=(-3,~8)$} \\ \text{Let $P_2(x_2,~y_2)=(8,~-5)$}\)

Formula: \(\text{distance} = \sqrt{ \left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

\(\begin{array}{|rcll|} \hline \mathbf{\text{distance}} &=& \mathbf{\sqrt{ \left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}} \\\\ \text{distance} &=& \sqrt{ \Big(8-(-3)\Big)^2+\left(-5-8\right)^2} \\\\ \text{distance} &=& \sqrt{ 11^2+\left(-13\right)^2} \\\\ \text{distance} &=& \sqrt{ 121+169} \\\\ \text{distance} &=& \sqrt{ 290} \\\\ \text{distance} &=& 17.0293863659 \\ \hline \end{array} \)

The distance is 17.0

Thank you Melody !

How many distinct diagonals of a convex heptagon (7-sided polygon) can be drawn?

\(6+5+4+3+2+1 - 7 = 14\)

Find the polynomial \(p(x)\), with real coefficients,
such that \(p(2) = 5\) and \(p(x) p(y) = p(x) + p(y) + p(xy) - 2\) for all real numbers \(x\) and \(y\).

Find all integers x for which there exists an integer y such that 1/x + 1/y = 1/7. In other words,

find all ordered pairs of integers (x,y) that satisfy this equation, then enter just the x's from these pairs.

In triangle ABC, angle A = 2 * angle B. 

Prove that the side lengths satisfy a^2 = b(b + c).

sin-rule

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{\sin(A)}{a}} &=& \mathbf{\dfrac{\sin(B)}{b}} \quad & | \quad A=2B \\\\ \dfrac{\sin(2B)}{a} &=& \dfrac{\sin(B)}{b} \quad & | \quad \sin(2B)=\sin(B)\cos(B) \\\\ \dfrac{\sin(B)\cos(B)}{a} &=& \dfrac{\sin(B)}{b} \quad & | \quad :\sin(B) \\\\ \dfrac{\cos(B)}{a} &=& \dfrac{1}{b} \\\\ \mathbf{\cos(B)} &=& \mathbf{\dfrac{a}{2b}} \\ \hline \end{array}\)

cos-rule

\(\begin{array}{|rcll|} \hline \mathbf{b^2} &=& \mathbf{a^2+c^2-2ac\cos(B)} \quad & | \quad \mathbf{\cos(B)=\dfrac{a}{2b}} \\ b^2 &=& a^2+c^2-2ac\dfrac{a}{2b} \\ b^2 &=& a^2+c^2-\dfrac{a^2c}{b} \\ \mathbf{\dfrac{a^2c}{b}} &=& \mathbf{a^2+c^2-b^2} \\ \hline \end{array} \)

cos-rule

\(\begin{array}{|rcll|} \hline \mathbf{a^2} &=& \mathbf{b^2+c^2-2bc\cos(2B)} \quad & | \quad \cos(2B)=\cos^2(B)-\sin^2(B) = 2\cos^2(B)-1 \\ a^2 &=& b^2+c^2-2bc\left(2\cos^2(B)-1\right) \quad & | \quad \mathbf{\cos(B)=\dfrac{a}{2b}} \\ a^2 &=& b^2+c^2-2bc\left(2\dfrac{a^2}{4b^2}-1\right) \quad & | \quad \mathbf{\cos(B)=\dfrac{a}{2b}} \\ a^2 &=& b^2+c^2-\dfrac{a^2c}{b}+2bc \quad & | \quad \mathbf{\dfrac{a^2c}{b}=a^2+c^2-b^2} \\ a^2 &=& b^2+c^2-(a^2+c^2-b^2)+2bc \\ a^2 &=& b^2+c^2-a^2-c^2+b^2+2bc \\ a^2 &=& b^2 -a^2 +b^2+2bc \\ 2a^2 &=& 2b^2 +2bc \quad & | \quad : 2 \\ a^2 &=& b^2 +bc \\ \mathbf{a^2} &=& \mathbf{b(b +c )} \\ \hline \end{array}\)

The shape ABC is an triangle, AB = 64 mi, BC = 32.6 mi. Angle ABC is 74 degrees.

Using the cosine rule, calculate the distance of AC

\(\begin{array}{|rcll|} \hline AC^2 &=& AB^2+BC^2-2*AB*BC*\cos(74^\circ) \\ AC^2 &=& 64^2+32.6^2-2*64*32.6*\cos(74^\circ) \\ AC^2 &=& 4096+1062.76-4172.8*\cos(74^\circ) \\ AC^2 &=& 4096+1062.76-4172.8*0.27563735582 \\ AC^2 &=& 5158.76-1150.17955835 \\ AC^2 &=& 4008.58044165 \\ \mathbf{AC} &=& \mathbf{63.3133512116} \\ \hline \end{array}\)

AC is 63.3 mi