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strojnjashka [21]
3 years ago
9

In the aftermath of a car accident, it is concluded that one driver slowed to a halt in 20 seconds while skidding 1775 feet. If

the speed limit was 60 miles per hour, can it be proved that the driver had been speeding? (Hint: 60 miles per hour is equal to 88 feet per second.) We can guarantee that at some time from when the driver first pressed on the brake to when the car came to a complete stop the car was traveling
Mathematics
1 answer:
-Dominant- [34]3 years ago
7 0
Assume uniformly accelerated motion

Vf = Vo - at => a= [Vo - Vf] / t

Vf = 0, t = 20 s => a = Vo / 20

x = Vot - at^2 / 2

x = Vot - (Vo / 20) * t^2 /2

x = 1775 feet * [1 mile / 5280 feet] = 0.336 mile

t = 20 s

=> 0.336 = 20Vo - Vo(20)^2 /(40) = 20Vo - 10 Vo = 10 Vo

=> Vo = 0.336 /10 = 0.0336 mile /s

Vo = 0.0336 mile/s * 3600 s/h = 120.96 mile / h

Therefore, the car was way over the speed limit.



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vesna_86 [32]

Answer:

22

Step-by-step explanation:

To solve this we can plug the values in.

-2x-3y+z

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8 0
3 years ago
Show work please<br> \sqrt(x+12)-\sqrt(2x+1)=1
Nesterboy [21]

Answer:

x=4

Step-by-step explanation:

Given \displaystyle\\\sqrt{x+12}-\sqrt{2x+1}=1, start by squaring both sides to work towards isolating x:

\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2

Recall (a-b)^2=a^2-2ab+b^2 and \sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b}:

\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2\\\implies x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1

Isolate the radical:

\displaystyle\\x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1\\\implies -2\sqrt{(x+12)(2x+1)}=-3x-12\\\implies \sqrt{(x+12)(2x+1)}=\frac{-3x-12}{-2}

Square both sides:

\displaystyle\\(x+12)(2x+1)=\left(\frac{-3x-12}{-2}\right)^2

Expand using FOIL and (a+b)^2=a^2+2ab+b^2:

\displaystyle\\2x^2+25x+12=\frac{9}{4}x^2+18x+36

Move everything to one side to get a quadratic:

\displaystyle-\frac{1}{4}x^2+7x-24=0

Solving using the quadratic formula:

A quadratic in ax^2+bx+c has real solutions \displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}. In \displaystyle-\frac{1}{4}x^2+7x-24, assign values:

\displaystyle \\a=-\frac{1}{4}\\b=7\\c=-24

Solving yields:

\displaystyle\\x=\frac{-7\pm \sqrt{7^2-4\left(-\frac{1}{4}\right)\left(-24\right)}}{2\left(-\frac{1}{4}\right)}\\\\x=\frac{-7\pm \sqrt{25}}{-\frac{1}{2}}\\\\\begin{cases}x=\frac{-7+5}{-0.5}=\frac{-2}{-0.5}=\boxed{4}\\x=\frac{-7-5}{-0.5}=\frac{-12}{-0.5}=24 \:(\text{Extraneous})\end{cases}

Only x=4 works when plugged in the original equation. Therefore, x=24 is extraneous and the only solution is \boxed{x=4}

4 0
2 years ago
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7 0
3 years ago
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