Draw two different shapes with a perimeter of 16 units. What is the area of each shape?

2 answers:

Nadusha1986 [10]3 years ago

8 0

<u>Shape #1: A square.</u>
Every side is 4 units long.
The perimeter is 16 units.
The area is <em>16 </em>square units.

<u>Shape #2: A rectangle.</u>
The length is 7.9 units.
The width is 0.1 unit.
The perimeter is 16 units.
The area is <em>0.79</em> of a square unit.

<u>Shape #3: A circle.</u>
The diameter is (16/π) units. (about 5.093 units)
The perimeter (circumference) is 16 units.
The area is (64/π) square units. (about <em>20.37</em> square units)

frosja888 [35]3 years ago

5 0

So, you will obviously have to do the drawing bit, but I can help with the rest. I would first recommend drawing a square with a side length of 4, which will give you a perimeter of 16. The formula for area in a square is s², where s is the side length. That means the area for the square is 4², or 16 units², just like the perimeter.

You can draw any shape you like for the second one, but I would suggest a rectangle with a length of 6 and a width of 2. Perimeter for a rectangle is P = 2L + 2w , and 2(6) +2(2) = 16. The area would then be the length times the width, or 2(6), which equals 12 units² for the second figure.

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Suppose you just received a shipment of nine televisions. Three of the televisions are defective. If two televisions are randoml

Temka [501]

<h2>Answer:</h2>

The probability that both televisions work is: 0.42

The probability at least one of the two televisions does not work is:

0.5833

<h2>Step-by-step explanation:</h2>

There are a total of 9 televisions.

It is given that:

Three of the televisions are defective.

This means that the number of televisions which are non-defective are:

9-3=6

The probability that both televisions work is calculated by:

$=\dfrac{6_C_2}{9_C_2}$

( Since 6 televisions are in working conditions and out of these 6 2 are to be selected.

and the total outcome is the selection of 2 televisions from a total of 9 televisions)

Hence, we get:

$=\dfrac{\dfrac{6!}{2!\times (6-2)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{\dfrac{6!}{2!\times 4!}}{\dfrac{9!}{2!\times 7!}}\\\\\\=\dfrac{5}{12}\\\\\\=0.42$

The probability at least one of the two televisions does not work:

Is equal to the probability that one does not work+probability both do not work.

Probability one does not work is calculated by:

$=\dfrac{3_C_1\times 6_C_1}{9_C_2}\\\\\\=\dfrac{\dfrac{3!}{1!\times (3-1)!}\times \dfrac{6!}{1!\times (6-1)!}}{\dfrac{9!}{2!\times (9-2)!}}\\\\\\=\dfrac{3\times 6}{36}\\\\\\=\dfrac{1}{2}\\\\\\=0.5$

and the probability both do not work is:

$=\dfrac{3_C_2}{9_C_2}\\\\\\=\dfrac{1}{12}\\\\\\=0.0833$

Hence, Probability that atleast does not work is:

0.5+0.0833=0.5833

4 0

4 years ago

CAN SOMEONE HELP EXPLAIN THIS TO ME !!!!

Jet001 [13]

The quadratic formula is above.

You want each equation in standard form: ax^2 + bx + c = 0
I begin each problem by defining variables.
For instance 3. Is in standard form. X^2 -2x - 3 = 0. a = 1, b = -2, c = -3

Now use quadratic formula: x = [- b + or - sqrt(b^2 - 4ac)]\2a