A room is 45 m x 15 m. What is the least number of rectangle tiles required to pave the floor of the room, if each tile is 50 cm x 1m?

Answers

Answer 1

Answer: Hey Batala426,

If the room is 45m x 15m, then the total area is 675 metres squared.
Each tile is 50cm x 1m, so the area of each one is 0.5 metres squared.

675 / 0.5 = 1350

So the least number of tiles you need to pave the whole room is 1350 tiles.

Hope this helps!

Chen :D

Answer with explanation:

Ques 1)

$x=(1)/(√(3)-√(2))$

Now we are asked to find the value of:

$√(x)-(1)/(√(x))$

We know that:

$(√(x)-(1)/(√(x)))^2=x+(1)/(x)-2$

Also:

$x=(1)/(√(3)-√(2))$ could be written as:

$x=(1)/(√(3)-√(2))* (√(3)+√(2))/(√(3)+√(2))\n\n\nx=(√(3)+√(2))/((√(3))^2-(√(2))^2)$

since, we know that:

$(a+b)(a-b)=a^2-b^2$

Hence,

$x=(√(3)+√(2))/(3-2)\n\n\nx=√(3)+√(2)$

Also,

$(1)/(x)=√(3)-√(2)$

Hence, we get:

$(√(x)-(1)/(√(x)))^2=√(3)+√(2)+√(3)-√(2)-2\n\n\n(√(x)-(1)/(√(x)))^2=2√(3)-2\n\n\n√(x)-(1)/(√(x))=\sqrt{2√(3)-2}$

Hence,

$√(x)-(1)/(√(x))=\sqrt{2√(3)-2}$

Ques 2)

$x=(√(a+2b)+√(a-2b))/(√(a+2b)-√(a-2b))$

on multiplying and dividing by conjugate of denominator we get:

$x=(√(a+2b)+√(a-2b))/(√(a+2b)-√(a-2b))* (√(a+2b)+√(a-2b))/(√(a+2b)+√(a-2b))\n\n\nx=((√(a+2b)+√(a-2b))^2)/((√(a+2b))^2-(√(a-2b))^2)\n\n\nx=((√(a+2b))^2+(√(a-2b))^2+2√(a+2b)√(a-2b))/(a+2b-a+2b)\n\n\nx=(a+2b+a-2b+2√(a+2b)√(a-2b))/(4b)\n\n\nx=(2a+2√(a^2-4b^2))/(4b)\n\n\nx^2=((2a+2√(a^2-4b^2))/(4b))^2\n\n\nx^2=((2a+2√(a^2-4b^2))^2)/(16b^2)$

Hence, we have:

$x^2=(4a^2+4(a^2-4b^2)+8a√(a^2-4b^2))/(16b^2)\n\n\nx^2=(4a^2+4a^2-16b^2+8a√(a^2-4b^2))/(16b^2)\n\n\n\nx^2=(8a^2-16b^2+8a√(a^2-4b^2))/(16b^2)\n\n\nbx^2=(8a^2-16b^2+8a√(a^2-4b^2))/(16b)\n\n\nbx^2=(8a(a+√(a^2-4b^2))-16b^2)/(16b)\n\n\nbx^2=(8a(a+√(a^2-4b^2)))/(16b)-(16b^2)/(16b)\n\n\nbx^2=(a(a+√(a^2-4b^2)))/(2b)-b\n\n\nbx^2=ax-b\n\n\ni.e.\n\n\nbx^2-ax+b=0$

1. x = 1/ ( $√(3) - √(2))$ = $√(3)+ √(2)$ ;
( $√(x) -1/ √(x) )^(2) = x + 1/x - 2 =$ =

A movie was shown 6 times every day from November 24 through December 19. How many times was the movie shown in all? (Note: The month of November has 30 days.)

Answers

Answer:

156 times

Step-by-step explanation:

Given: A movie was shown 6 times every day from November 24 through December 19.

To find: The number of times the movie shown in all.

Solution:

To find the number of times the movie shown in all, we need to first find the number of days the movie is shown.

So, we have to find the number of days from November 24 through December 19.

Number of days the movie is shown in November is 7 days.

In December the movie is shown for 19 days.

So, we get the movie is shown for 26 days in all.

It is given that, each day the movie is shown 6 times.

So, to find the total number of times the movie is shown we need to multiply 26 and 6.

Therefore, we get

number of times movie shown $=26*6=156$

Hence, the movie is shown 156 times.

Answer:

Answer: A. 138 times

How?

During that time period there were 23 days so 23 multiplied by the amount of times played per day (6)

You get the answer 138

Step-by-step explanation:

If 3 1/2 pounds of bananas cost $.98, how much would one pound cost?

Answers

The way you would find how much one pound costs is divide .98 by 3 1/2. 3 1/2 in decimal form is 3.5 so .98 ÷ 3.5 = .28

So one pound of bananas would cost $0.28

Hope that helped!

98/3.5=28 centsratio and proportion3.5/98=1/xcross multiplyx= 28 cents

Determine whether each relation represents a function. For each function, state the domain and range. 1) {(2,6), (-3,6), (4,9), (1,10)}
2) {(1,3), (2,3), (3,3), (4,3)}
3) {(-2,4), (-2,6), (0,3), (3,7)}
4) {(-2,4), (-1,1), (0,0), (1,1)}

Answers

Option 1 , 2 and 4 is defined as a function.

And, for 1 relation;

The domain of relation = {2, -3, 4, 1}

And, Range of relation = {6, 9 ,10}

For 2 relation;

The domain of relation= {1, 2, 3, 4}

The range of relation = {3}

For 4 relation ;

The domain of the relation = {-2, -1, 0, 1}

The range of the relation = {4, 1, 0}

What is function?

A function is defined as a relation between a set of input having only one output.

Now,

The relation is;

1) {(2,6), (-3,6), (4,9), (1,10)}

Hence, The domain of relation = {2, -3, 4, 1}

And, Range of relation = {6, 9 ,10}

Clearly, Each input (domain) has only one output (range)

Hence, The given relation is function.

For,The relation;

2) {(1,3), (2,3), (3,3), (4,3)}

The domain of relation= {1, 2, 3, 4}

The range of relation = {3}

Clearly, Each input (domain) has only one output (range).

Hence, The given relation is function.

For, The function;

3) {(-2,4), (-2,6), (0,3), (3,7)}

Clearly, -2 has two image 4 and 6.

So, This relation is not a function.

For, The function;

4) {(-2,4), (-1,1), (0,0), (1,1)}

The domain of the relation = {-2, -1, 0, 1}

The range of the relation = {4, 1, 0}

Clearly, Each input (domain) has only one output (range).

Hence, This relation is a function.

Therefore,

Option 1 , 2 and 4 is defined as a function.

And, for 1 relation;

The domain of relation = {2, -3, 4, 1}

And, Range of relation = {6, 9 ,10}

For 2 relation;

The domain of relation= {1, 2, 3, 4}

The range of relation = {3}

For 4 relation ;

The domain of the relation = {-2, -1, 0, 1}

The range of the relation = {4, 1, 0}

Learn more about the function visit:

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$D-domain;\ R-range\n\n1)\ \{(2;\ 6),\ (-3;\ 6),\ (4;\ 9),\ (1;\ 10)\}\ YES\ :)\nD=\{2;-3;\ 4;\ 1\};\ R=\{6;\ 9;\ 10\}\n\n2)\ \{(1;\ 3),\ (2;\ 3),\ (3;\ 3),\ (4;\ 3)\}\ \ \ YES\ :)\nD=\{1;\ 2;\ 3;\ 4\};\ R=\{3\}\n\n3)\ \{(\fbox{-2};\ 4),\ (\fbox{-2};\ 6),\ (0;\ 3),\ (3;\ 7)\}\ \ \ NO\ :(\n\n4)\ \{(-2;\ 4),\ (-1;\ 1),\ (0;\ 0),\ (1;\ 1)\}\ \ \ \ YES\ :)\nD=\{-2;-1;\ 0;\ 1\};\ R=\{4;\ 1;\ 0\}$

The lateral area of a cube is 36 square inches. How long is each edge?

Answers

The length of the edge of the given cube that has a lateral area of 36 square inches is: 3 inches.

What is the Lateral Area of a Cube?

Lateral area of a cube = 4a², wherea is the length of each edge of the cube.

Given the parameters:

Lateral area = 36 square inches
Length of each edge = a = ?

4a² = 36

a² = 36/4

a² = 9

a = 3 inches.

Length of each edge of the cube = 3 inches.

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Cubes have three square sides. The area of one side is the edge length squared:
36 = L^2
L = sqrt(36) = 6
The volume of a cube is edge length cubed:
V = L^3 = 216 cubic inches

Bc reflects about a line such that N is the reflection of B and O is the reflection of C. Point N is shown on the coordinate plane, but point O is not. What will the coordinates of point O be?

Answers

From my research, the coordinate points can be found on image below. Since N is the reflection of B and O is the reflection of C, if we were to draw a line to the right of N and a vertical line below C, the point of intersection of these two lines should give us the location of point O. Doing just that, point O is found at (5,5).

The coordinates of point O of the given Reflection is; 5, 5

How to find the point coordinates?

The coordinates point is missing and so I have attached it.

Now, from the question, we are told that N is the reflection of B and O is the reflection of C. Thus, from the attached image, drawing a line to the right of N and a vertical line below C, the point of intersection of these two lines would give us the location of point O.

Drawing a line to the right of N and a vertical line below C, the point of intersection of these two lines which is point O has coordinates at (5,5).

Read more about Point Coordinates at; brainly.com/question/17206319

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