Idon't know how to find this specific question, i'm so confused.


Idon't know how to find this specific question, i'm so confused.

Answers

Your total will be $88.15 and this helps and good luck
Two similar triangles would be Triangle XBY and triangle ABC. You have to look at the small and large aspects of a diagram.
The first triangle is the entire figure except for the line running across the middle (triangle ABC). The second triangle is created with the line running across the middle because it forms a mini one at the top half of triangle ABC.  

Your two triangles are triangle ABC and triangle XBY. 

Hope this helps!! :)

recalling that d = rt, distance = rate * time.


we know Hector is going at 12 mph, and he has already covered 18 miles, how long has he been biking already?


\bf \begin{array}{ccll} miles&hours\\ \cline{1-2} 12&1\\ 18&x \end{array}\implies \cfrac{12}{18}=\cfrac{1}{x}\implies 12x=18\implies x=\cfrac{18}{12}\implies x=\cfrac{3}{2}


so Hector has been biking for those 18 miles for 3/2 of an hour, namely and hour and a half already.

then Wanda kicks in, rolling like a lightning at 16mph.

let's say the "meet" at the same distance "d" at "t" hours after Wanda entered, so that means that Wanda has been traveling for "t" hours, but Hector has been traveling for "t + (3/2)" because he had been biking before Wanda.

the distance both have travelled is the same "d" miles, reason why they "meet", same distance.


\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ Hector&d&12&t+\frac{3}{2}\\[1em] Wanda&d&16&t \end{array}\qquad \implies \begin{cases} \boxed{d}=(12)\left( t+\frac{3}{2} \right)\\[1em] d=(16)(t) \end{cases}


\bf \stackrel{\textit{substituting \underline{d} in the 2nd equation}}{\boxed{(12)\left( t+\frac{3}{2} \right)}=16t}\implies 12t+18=16t \\\\\\ 18=4t\implies \cfrac{18}{4}=t\implies \cfrac{9}{2}=t\implies \stackrel{\textit{four and a half hours}}{4\frac{1}{2}=t}

recalling that d = rt, distance = rate * time.


we know Hector is going at 12 mph, and he has already covered 18 miles, how long has he been biking already?


\bf \begin{array}{ccll} miles&hours\\ \cline{1-2} 12&1\\ 18&x \end{array}\implies \cfrac{12}{18}=\cfrac{1}{x}\implies 12x=18\implies x=\cfrac{18}{12}\implies x=\cfrac{3}{2}


so Hector has been biking for those 18 miles for 3/2 of an hour, namely and hour and a half already.

then Wanda kicks in, rolling like a lightning at 16mph.

let's say the "meet" at the same distance "d" at "t" hours after Wanda entered, so that means that Wanda has been traveling for "t" hours, but Hector has been traveling for "t + (3/2)" because he had been biking before Wanda.

the distance both have travelled is the same "d" miles, reason why they "meet", same distance.


\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ Hector&d&12&t+\frac{3}{2}\\[1em] Wanda&d&16&t \end{array}\qquad \implies \begin{cases} \boxed{d}=(12)\left( t+\frac{3}{2} \right)\\[1em] d=(16)(t) \end{cases}


\bf \stackrel{\textit{substituting \underline{d} in the 2nd equation}}{\boxed{(12)\left( t+\frac{3}{2} \right)}=16t}\implies 12t+18=16t \\\\\\ 18=4t\implies \cfrac{18}{4}=t\implies \cfrac{9}{2}=t\implies \stackrel{\textit{four and a half hours}}{4\frac{1}{2}=t}

  \bold{c)\quad x=\pm \dfrac{\sqrt{26}}{2}\qquad \qquad d)\quad x=\pm 2\sqrt3}

Step-by-step explanation:

It is much easier to use the square root method but since you requested the quadratic formula, I will solve it using that method.

First, replace g(x) with 0 for letter c and 11 for letter d. Then simplify and move everything to one side so the equation is equal to zero.  Then plug in the a, b, and c values into the quadratic formula to solve for x.

g(x) + 5 = 2x² - 8

c) 0 + 5 = 2x² - 8

    0       = 2x² - 13     (subtracted 5 from both sides)

-->      0 = 2x² + 0x - 13         --> a = 2, b = 0, c = -13

x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-(0)\pm \sqrt{(0)^2-4(2)(-13)}}{2(2)}\\\\\\.\quad =\pm\dfrac{\sqrt{104}}{4}\\\\\\.\quad =\pm \dfrac{2\sqrt{26}}{4}\\\\\\.\quad =\large\boxed{\pm \dfrac{\sqrt{26}}{2}}

d) 11 + 5 = 2x² - 8

    0       = 2x² - 24     (subtracted 16 from both sides)

-->      0 = 2x² + 0x - 24         --> a = 2, b = 0, c = -24

x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-(0)\pm \sqrt{(0)^2-4(2)(-24)}}{2(2)}\\\\\\.\quad =\pm\dfrac{\sqrt{192}}{4}\\\\\\.\quad =\pm \dfrac{8\sqrt{3}}{4}\\\\\\.\quad =\large\boxed{\pm 2\sqrt3}

The answer is D because all you have to do is subtract the two numbers in the water tank.
The answer is D because all you have to do is subtract the two numbers in the water tank.


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