Someone please help me

Can someone please help me with this ?

Galina-37 [17]

Answer:

Step-by-step explanation:

<h3>Haiaisiisisisisiisjsjsjsjjjxjxixijsbsbs</h3>

3 0

3 years ago

Whats 1/5 - 1/7 please can you tell me how to work it out<br> thanks

Andrej [43]

Convert the fractions so they both have the same denominator (bottom part of the fraction)

1/5 = 7/35

This is because 5*7=35, so 1*7=7

1/7 = 5/35

This is because 7*5=35, so 1*5 =5

Then subtract the top numbers to get the answer:

$\frac{7}{35}-\frac{5}{35} = \frac{2}{35}$

3 0

4 years ago

Read 2 more answers

Ten is five lees than three times a number. Write as an algebraic equation.

shutvik [7]

Is means equals. The number before the words Less than means that number is subtracted from that number. 3 times a number, we don’t know the number so we use a variable, 3x. So..........

3x-5=10

5 0

4 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

You are planning to retile the walls around your bathtub. You measure the dimensions of the wall and find that the length of the

blsea [12.9K]

Answer:

it is the best deal cause i said

Step-by-step explanation:

8 0

3 years ago