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nalin [4]
4 years ago
11

Four women must cross a bridge over a deep ravine in enemy territory in the middle of the night, treacherous bridge will hold on

two women at once and it's necessary to carry a lantern while crossing. (Person must stay together while crossing but may return alone) . One of the woman requires five mins for the trip across, one takes ten minu, a third requires 20 and the last takes 25 minutes (each of them can walk slower if necessary but no faster). Unfortunately they only have one lantern among them. How can they make the cro
Mathematics
1 answer:
Galina-37 [17]4 years ago
8 0
Ok, this is usually a trial and error question and it is supposed to be fun.
I believe you want the four women to cross the bridge in the least possible time before the bridge collapses.

The solution is as follows:
<span>1- The 5-minute lady and 10-minute lady cross the bridge
    (the total time will be 10 minutes)
</span>2- <span>The 5-minute lady returns
    (the total time will be 5 + 10 = 15 minutes)
3- </span><span>The 20-minute lady and 25-minute lady cross
    (the total time will be 15 + 25 = 40 minutes)
4- </span><span>The 10-minute lady returns 
    (the total time will be 10 + 40 = 50 minutes)
5- </span><span>The 5-minute lady and 10-minute lady finally cross
    (the total time will be 10 + 50 = 60 minutes)

The last one to cross should step aside from the bridge as quickly as possible before the bridge collapses.</span>
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Is the square root of 900 rational
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Answer:

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150 % of 60 is what number?
beks73 [17]

Answer:

90

Step-by-step explanation:

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3 years ago
One hundred times the quantity eight plus five given expression symbols
Lady_Fox [76]
The answer is 1100 because it is (8 + 5) x 100
7 0
3 years ago
Carlos is using the quadratic formula to find the solutions of y=3x^2-5x-2. Which of the following will simplify to the correct
Nana76 [90]

Answer:

Fx=\frac{5\pm\sqrt{25+24} }{6}

Step-by-step explanation:

A quadratic equation in one variable given by the general expression:

ax^2+bx+c

Where:

a\neq 0

The roots of this equation can be found using the quadratic formula, which is given by:

x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}

So:

y(x)=3x^2-5x-2=0

As you can see, in this case:

a=3\\b=-5\\c=-2

Using the quadratic formula:

x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}=\frac{-(-5)\pm\sqrt{(-5)^2-4(3)(-2)} }{2(3)}=\frac{5\pm\sqrt{25+24} }{6}

Therefore, the answer is:

Fx=\frac{5\pm\sqrt{25+24} }{6}

4 0
3 years ago
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