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neonofarm [45]
3 years ago
11

A square room is 15 m long, find the are of floor​

Mathematics
2 answers:
Snowcat [4.5K]3 years ago
7 0

Answer:

225 square meters

Step-by-step explanation:

Area of a square:

A = w², where w is the width of the square

Plug in values:

A = 15m²

A = 225 square meters

DiKsa [7]3 years ago
7 0

Answer:

A = 225m^{2}

Step-by-step explanation:

The area of a square is;

A = l^{2}

A = Length * Length

A = 15m^{2}

A = 15m * 15m

A = 225m^{2}

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She can have 11 rows with 3 in the last row and she can have 6 rows with 3 in the last row
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3 years ago
A rectangle has a length 5 meters more than five times the
k0ka [10]

Given:

A rectangle has a length 5 meters more than five times the  width.

The area of the rectangle is less than 100 meters  squared.

To find:

The expression or inequality that represents all possible  widths of the rectangle.

Solution:

Let x be the width of the rectangle.

Length of the rectangle is 5 meters more than five times the  width.

length=5x+5

Area of rectangle is

Area=length \times width

Area=(5x+5) \times x

Area=5x^2+5x

The area of the rectangle is less than 100 meters  squared.

5x^2+5x

5x^2+5x-100

Divide both sides by 5.

x^2+x-20

x^2+5x-4x-20

x(x+5)-4(x+5)

(x+5)(x-4)

It is true if one factor is negative and other is positive. So,

x-4     ...(i)

x+5>0\Rightarrow x>-5       ...(ii)

Using (i) and (ii), we get

-5

Therefore, the required expression or inequality for possible

widths of the rectangle is -5.

4 0
3 years ago
How many 1/3 cups are in 2/3 cup?
KengaRu [80]

Answer: 2 , 1/3 cups

Step-by-step explanation:

1/3 is the base number and 2/3 is another third being added onto the singular 1/3. The awnser would be 2 because 1/3+1/3= 2/3 or 2/3/2= 1/3

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Gnesinka [82]
(-3, -3) and (2, 7).

The solutions to a graphed system of equations are where the two graphs cross each other. So in this case the two equations cross at (-3, -3) and (2, 7), and those points are the solutions to the system of equations.
7 0
3 years ago
Describe a transformation or series of transformations that demonstrates the congruence between figures A and B.
N76 [4]

Answer:

90 clockwise (or counterclockwise) rotation and then a reflection over the axis between the two shape (those two steps go in any order)

Step-by-step explanation:

for this lets mark the innermost point of each shape a (blue or A) and a' (red or B)* and the second point b and b'

here we see that the two shapes are in a position to where they seem reflected over a non-existent third diagonal axis, though this is not the case, we need to bring the shape into a position where it can be transformed to the quadrant of shape B and overlap the shape

so when you have a reflection over a diagonal axis, we can rotate or reflect the shape to a new quadrant, and perform the step thats not the first, so say we made a reflection over the X-axis, the shape is now in the lower half of the graph with shape B, from here we perform our last step wich is to rotate the shape into the quadrant of shape B in a clockwise motion, now a and a' overlap and b and b' overlap, same for c, c',d and d'

(*the ' in this case is called a prime symbol, when used, distinguishes two points or lines on a graph, A' = A prime)

5 0
3 years ago
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