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3 years ago

13

A tap takes 3 hours to fill a tank. A discharge tap takes 6 hours to empty the same tank. How long will it take to fill the tank

with both taps open?

1 answer:

wlad13 [49]3 years ago

8 0

Answer:

2 hours

Step-by-step explanation:

first tap= 3hours

therefore work done is = 1/3

second tap= 6hours

therefore work done = 1/6

so, 1/3+ 1/6

=2/6 + 1/6

=3/6

=1/2

=2/1 hours

=2 hours

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Find the 6th term of the expansion of (2p - 3q)11. a. -7,185,024p4q7 c. -7,185p4q7 b. -7,185,024p6q5 d. -7,185p6q5

bezimeni [28]

$\bf (2p-3q)^{11}\implies 
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&&(2p)^{11}(-3q)^0\\
2&+11&(2p)^{10}(-3q)^1\\
3&+55&(2p)^9(-3q)^2\\
4&+165&(2p)^8(-3q)^3\\
5&+330&(2p)^7(-3q)^4\\
6&+462&(2p)^6(-3q)^5
\end{array}$

the coefficient for the first term is 1, the next is 11 and so on... now, notice, the elements of the binomial, the 1st element starts off with 11, and every term it goes down by 1, the 2nd element starts off at 0, and goes up by 1 in each term.

now, to get the next coefficient, you simply, "get the product of the current coefficient and the exponent of the 1st element, and divide that by the exponent of the 2nd element in the next term".

for example, how did we get 165 for the 4th term.... well (55*9)/3

how did we get 462 for the 6th term? well (330*7)/5.

and then you can just expand it from there.

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Allushta [10]

Check for critical points within the unit ball by solving for when the first-order partial derivatives vanish:
$f_x=5yz=0\implies y=0\text{ or }z=0$
$f_y=5xz=0\implies x=0\text{ or }z=0$
$f_z=5xy=0\implies x=0\text{ or }y=0$

Taken together, we find that (0, 0, 0) appears to be the only critical point on $f$ within the ball. At this point, we have $f(0,0,0)=0$ .

Now let's use the method of Lagrange multipliers to look for critical points on the boundary. We have the Lagrangian

$L(x,y,z,\lambda)=5xyz+\lambda(x^2+y^2+z^2-1)$

with partial derivatives (set to 0)

$L_x=5yz+2\lambda x=0$
$L_y=5xz+2\lambda y=0$
$L_z=5xy+2\lambda z=0$
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We then observe that

$xL_x+yL_y+zL_z=0\implies15xyz+2\lambda=0\implies\lambda=-\dfrac{15xyz}2$

So, ignoring the critical point we've already found at (0, 0, 0),

$5yz+2\left(-\dfrac{15xyz}2\right)x=0\implies5yz(1-3x^2)=0\implies x=\pm\dfrac1{\sqrt3}$
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Answer:

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