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3 years ago

Find the value of L. Perimeter = 20. Width is 6. So what is L? On a square

2 answers:

8 0

Well first of all, that's not a square. On a square, all 4 sides are
the same, and the length is the same distance as the width. If the
width of a square is 6, and all four sides are the same, then the
perimeter is (4 x 6) = 24.

So the shape in your question may be a rectangle, but it's definitely
not a square one.

The perimeter is the distance an ant has to walk if you put him down
on one corner and he walks all the way around the shape until he gets
back to where he started from.
That's the perimeter.

If you set the ant down on the corner of a rectangle, then he walks the
length, then the width, the the length again, then the width again, and
he's back where he started.

That was 2 lengths and 2 widths ... the perimeter of every rectangle.

Take 2 widths away from the perimeter, and you have 2 lengths left. OK ?

You said that the width of your rectangle is 6. Two widths is 12.
Take the 2 widths away from the perimeter, and you're left with (20 - 12) = 8.
But that's 2 lengths !
So the length is half of 8 = 4 .

professor190 [17]3 years ago

5 0

The easiest way to tell you the answer is, The answer is L=4.

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A 5 mile jogging path diagonal divides a rectangular park in half. The park is 4 miles long. Find the width of the park.

levacccp [35]

Answer: The required width of the park is 3 miles.

Step-by-step explanation: Given that a 5 mile jogging path diagonal divides a rectangular park in half and the park is 4 miles long.

We are to find the width of the park.

As shown in the attached figure below, the diagonal BD divides the rectangle ABCD into two equal parts,

where length DA = 5 miles, BD = 4 miles and AB = width = ?

Since each angle of a rectangle is right-angle, so triangle ABD will be a right-angled triangle.

Using Pythagoras theorem in triangle ABD, we have

$BD^2=AB^2+DA^2\\\\\Rightarrow AB=\sqrt{BD^2-DA^2}\\\\\Rightarrow AB=\sqrt{5^2-4^2}\\\\\Rightarrow AB=\sqrt{25-16}\\\\\Rightarrow AB=\sqrt{9}\\\\\Rightarrow AB=3.$

Thus, the required width of the park is 3 miles.

7 0

3 years ago

Find the surface area of the figure below. Round your answer to the nearest hundredth if necessary,

olya-2409 [2.1K]

A= 2(w l + h l + h w) = 224m^2

7 0

3 years ago

The value of

marusya05 [52]

$\large\underline{\sf{Solution-}}$

We have to find out the value of the fraction.

Let us assume that:

$\sf \longmapsto x =2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + ... \infty} } }$

We can also write it as:

$\sf \longmapsto x =2 + \dfrac{1}{x}$

$\sf \longmapsto x =\dfrac{2x + 1}{x}$

$\sf \longmapsto {x}^{2} =2x + 1$

$\sf \longmapsto {x}^{2} - 2x - 1 = 0$

Comparing the given equation with ax² + bx + c = 0, we get:

$\sf \longmapsto\begin{cases} \sf a =1 \\ \sf b = - 2 \\ \sf c = - 1 \end{cases}$

By quadratic formula:

$\sf \longmapsto x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

$\sf \longmapsto x = \dfrac{2 \pm \sqrt{ {( - 2)}^{2} - 4(1)( - 1)} }{2 \times 1}$

$\sf \longmapsto x = \dfrac{2 \pm \sqrt{4 + 4} }{2 \times 1}$

$\sf \longmapsto x = \dfrac{2 \pm \sqrt{8} }{2}$

$\sf \longmapsto x = \dfrac{2 \pm2 \sqrt{2} }{2}$

$\sf \longmapsto x = 1 \pm\sqrt{2}$

$\sf \longmapsto x = \begin{cases} \sf 1 + \sqrt{2} \\ \sf 1 - \sqrt{2} \end{cases}$

But "x" cannot be negative. Therefore:

$\sf :\implies x = 1 + \sqrt{2}$

So, the value of the fraction is 1 + √2.

4 0

2 years ago

2 1/2 ÷ 1 5/8 = Two and a half divided by one and five eights

LUCKY_DIMON [66]

So first of all we can't start by changing them both into improper fractions

So it would be:
5/2 divided by 13/8
From there you would do 5/2 * 8/13 which could convert to
5/1 * 4/13 which would be 20/13

3 0

3 years ago

Read 2 more answers

The area of a rug is represented by the following expression, z2 + 4x – 12 , which of the following expressions

ch4aika [34]

Answer:

Width = (x-2)

Step-by-step explanation:

The area of a rug is given by $(x^2+4x-12)$

The length of the rug is (x+6)

We need to find the width of the rug.

The area of a rug is given by :

$A=lb\\\\x^2+4x-12=(x+6)b\\\\b=\dfrac{x^2+4x-12}{(x+6)}\\\\b=\dfrac{x^2+6x-2x-12}{(x+6)}\\\\b=\dfrac{x(x+6)-2(x+6)}{(x+6)}\\\\b=\dfrac{(x+6)(x-2)}{(x+6)}\\\\b=(x-2)$

So, the width of the rug is (x-2). Hence, the correct option is (c).

3 0

3 years ago