All categories

3 years ago

What is the area of the figure

Mathematics

1 answer:

Juli2301 [7.4K]3 years ago

5 0

The area of a figure is the number of squares required to cover it completely, like tiles on a floor. Area of a square = side times side. Since each side of a square is the same, it can simply be the length of one side squared.

You might be interested in

How long does it take the snail to crawl 82 inches? Enter your answer in a decimal number.

Sholpan [36]

Answer:

2827.5

Step-by-step explanation:

You have to divide 82 inches (distance) by the speed of the snail which is 0.029mph then you get the answer above.

8 0

3 years ago

A. Equilateral B. Right C. Obtuse D. Acute E. Scalene

faltersainse [42]

Answer:

A) Scalene E) Obtuse

Detailed Explanation:

A) Scalene

This triangle would be classified as scalene because none of the angles are equal.

____________________________________________________________

E) Obtuse

This triangle would be classified as an scalene triangle because one of the angles is larger than 90 degrees.

8 0

3 years ago

Read 2 more answers

What is 1.5 Devided by 1

VARVARA [1.3K]

Answer:

1.5

Step-by-step explanation:

Anything divided by 1 is itself

6 0

3 years ago

Read 2 more answers

What is 5/7 + 3/7??????

vaieri [72.5K]

8/7 as an improper and 1 and 1/7 for proper.

6 0

3 years ago

Read 2 more answers

Without plotting points, let M=(-2,-1), N=(3,1), M'= (0,2), and N'=(5, 4). Without using the distanceformula, show that segments

kramer

Given:

M=(x1, y1)=(-2,-1),

N=(x2, y2)=(3,1),

M'=(x3, y3)= (0,2),

N'=(x4, y4)=(5, 4).

We can prove MN and M'N' have the same length by proving that the points form the vertices of a parallelogram.

For a parallelogram, opposite sides are equal

If we prove that the quadrilateral MNN'M' forms a parallellogram, then MN and M'N' will be the oppposite sides. So, we can prove that MN=M'N'.

To prove MNN'M' is a parallelogram, we have to first prove that two pairs of opposite sides are parallel,

Slope of MN= Slope of M'N'.

Slope of MM'=NN'.

$\begin{gathered} \text{Slope of MN=}\frac{y2-y1}{x2-x1} \\ =\frac{1-(-1)}{3-(-2)} \\ =\frac{2}{5} \\ \text{Slope of M'N'=}\frac{y4-y3}{x4-x3} \\ =\frac{4-2}{5-0} \\ =\frac{2}{5} \end{gathered}$

Hence, slope of MN=Slope of M'N' and therefore, MN parallel to M'N'

$\begin{gathered} \text{Slope of MM'=}\frac{y3-y1}{x3-x1} \\ =\frac{4-(-1)}{5-(-2)} \\ =\frac{3}{2} \\ \text{Slope of NN'=}\frac{y4-y2}{x4-x2} \\ =\frac{4-1}{5-3} \\ =\frac{3}{2} \end{gathered}$

Hence, slope of MM'=Slope of NN' nd therefore, MM' parallel to NN'.

Since both pairs of opposite sides of MNN'M' are parallel, MM'N'N is a parallelogram.

Since the opposite sides are of equal length in a parallelogram, it is proved that segments MN and M'N' have the same length.

7 0

1 year ago