Help me guys...:( prove that 0!=1

a dive ring at the bottom of the pool is 10 feet below the surface of the water. sabine dives down and brings the ring back to t

icang [17]

Answer:

-10 feet below the surface of the water

Step-by-step explanation:

We are told that the dive rings position at the bottom of the pool is 10 feet below the surface of the water.

Now, an integer is basically a whole number that can either be be zero, positive or negative.

In this case, it's below the surface which means it's below zero.

Thus, it's negative.

So the integer is -10 feet below the water surface.

7 0

3 years ago

Question 6

likoan [24]

The answer is B 20,250 cm

7 0

3 years ago

Answer this! 18 divides x

Darya [45]

this makes no sense so maybe try something better

3 0

2 years ago

Anyone can help with this ?

Reil [10]

Answer:

Reflection

Step-by-step explanation:

The shape A'B'C' is just reflected from the shape ABC.

Hope this helps :)

6 0

3 years ago

Read 2 more answers

A rectangular swimming pool is bordered by a concrete patio. the width of the patio is the same on every side. the area of the s

andre [41]

Answer:

$x = \frac{1}{4}\left(-(l + w) + \sqrt{l^2 + 6lw + w^2} \right)$

where

l = length of the pool (w/o the patio)
w = width of the pool (w/o the patio)

Explanation:

Let

x = width of the patio
l = length of the pool (w/o the patio)
w = width of the pool (w/o the patio)

Since the pool is bordered by a complete patio,

Length of the pool (with the patio)
= (length of the pool (w/o the patio)) + 2*(width of the patio)
Length of the pool (with the patio) = l + 2x

Width of the pool (with the patio)
= (width of the pool (w/o the patio)) + 2*(width of the patio)
Width of the pool (with the patio) = w + 2x

Note that

Area of the pool (w/o the patio)
= (length of the pool (w/o the patio))(width of the pool (w/o the patio))
Area of the pool (w/o the patio) = lw

Area of the pool (with the patio)
= (length of the pool (w/o the patio))(width of the pool (w/o the patio))
= (l + 2x)(w + 2x)
= w(l + 2x) + 2x(l + 2x)
= lw + 2xw + 2xl + 4x²
Area of the pool (with the patio) = 4x² + 2x(l + w) + lw

Area of the patio
= (Area of the pool (with the patio)) - (Area of the pool (w/o the patio))
= (4x² + 2x(l + w) + lw) - lw
Area of the patio = 4x² + 2x(l + w)

Since the area of the patio is equal to the area of the surface of the pool, the area of the patio is equal to the area of the pool without the patio. In terms of the equation,

Area of the patio = Area of the pool (w/o the patio)
4x² + 2x(l + w) = lw
4x² + 2x(l + w) - lw = 0 (1)

Let

a = numerical coefficient of x² = 4
b = numerical coefficient of x = 2(l + w)
c = constant term = -lw

Then using quadratic formula, the roots of the equation 4x² + 2x(l + w) - lw = 0 is given by

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\\ = \frac{-2(l + w) \pm \sqrt{(2(l + w))^2 - 4(4)(-lw)}}{2(4)} 
\\ = \frac{-2(l + w) \pm \sqrt{(4(l + w)^2) + 16lw}}{8} 
\\ = \frac{-2(l + w) \pm \sqrt{(4(l^2 + 2lw + w^2) + 4(4lw)}}{8}
\\ = \frac{-2(l + w) \pm \sqrt{(4(l^2 + 2lw + w^2 + 4lw)}}{8}
\\ = \frac{-2(l + w) \pm \sqrt{(4(l^2 + 6lw + w^2)}}{8}$
$= \frac{-2(l + w) \pm 2\sqrt{l^2 + 6lw + w^2}}{8} \\= \frac{2}{8}(-(l + w) \pm \sqrt{l^2 + 6lw + w^2}) \\x = \frac{1}{4}(-(l + w) \pm \sqrt{l^2 + 6lw + w^2}) \\\boxed{x = \frac{1}{4}\left(-(l + w) + \sqrt{l^2 + 6lw + w^2} \right) \text{ or }}
\\\boxed{x = -\frac{1}{4}\left((l + w) + \sqrt{l^2 + 6lw + w^2} \right)}$

Since $(l + w) + \sqrt{l^2 + 6lw + w^2} \ \textgreater \ 0$ , $-\frac{1}{4}\left((l + w) + \sqrt{l^2 + 6lw + w^2}\right)$ is negative. Since x represents the patio width, x cannot be negative. Hence, the patio width is given by

$\boxed{x = \frac{1}{4}\left(-(l + w) + \sqrt{l^2 + 6lw + w^2} \right)}$

7 0

3 years ago

Help me guys...:(prove that 0!=1​

Help me guys...:(prove that 0!=1