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3 years ago

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there are 3 good guys and three bad guys on one side of a river full of piranhas and there is a canoe big enough for two people

and the bad guys out number the good guys how do you get everyone across

1 answer:

labwork [276]3 years ago

4 0

If there are 3 bad guys and 3 good guys then they don't out number the good guys.

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Product in simplest form.<br> -blank y^2-{5y-y(-7-9)]-[-y(15y+4)]=0

solmaris [256]

Answer:

hope this helps

Step-by-step explanation:

PLZZZ MARK BRAINLIEST

4 0

3 years ago

What is the difference between 5/6 and 1/5

Slav-nsk [51]

Hey there!

The answer is $\frac{19}{30}$

To solve this we use subtraction, because "the difference" represents subtraction, so:

$\frac{5}{6} - \frac{1}{5}$

Now, we must find a common denominator, and we can see that $6$ and $5$ both go into $30$ . To get to $30$ , we multiplied the first fraction by 5 and the second by 6, so we now have:

$\frac{25}{30} - \frac{6}{30}$

This is equal to $\frac{19}{30}$

Hope it helps and have a great day!

7 0

3 years ago

Paul is making grey paint.

velikii [3]

Its quite simple. for every white bucket of paint he uses, he uses 3 buckets of black paint as well. so if he made 24 litres of grey paint then he would have had to use 24 buckets of white paint. but, this might be a trick question because its saying he made 24 "litres" but the unit it started with was "buckets" so it might be trying to trick you because we dont know how much paint is in one "bucket"

5 0

3 years ago

Consider the function g(x) = (x-e)^3e^-(x-e). Find all critical points and points of inflection (x, g(x)) of the function g.

Elden [556K]

Answer:

The answer is " $cirtical\ points \ (x,g(x))\equiv (e,0),(e+3,\frac{27}{e^3})$ "

Step-by-step explanation:

Given:

$g(x) = (x-e)^3e^{-(x-e)}$

Find critical points:

$g(x) = (x-e)^3e^{(e-x)}$

differentiate the value with respect of x:

$\to g'(x)= (x-e)^3 \frac{d}{dx}e^{e-r} +e^{e-r} \frac{d}{dx}(x-e)^3=(x-e)^2 e^{(e-x)} [-x+e+3]$

critical points $g'(x)=0$

$\to (x-e)^2 e^{(e-x)} [e+3-x]=0\\\\\to e^{(e-x)}\neq 0 \\\\\to (x-e)^2=0\\\\ \to [e+3-x]=0\\\\\to x=e\\\\\to x=e+3\\\\\to x= e,e+3$

So,

The critical points of $(x,g(x))\equiv (e,0),(e+3,\frac{27}{e^3})$

7 0

3 years ago

The difference of two numbers is less than the greater number

kirill115 [55]

Sorry, i dont understand...

3 0

3 years ago

Read 2 more answers