1+3+5......+99 Si rezolvarea va rog fiindca nu inteleg !!

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Answer 1

Answer: I hope this helps you

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a community hall is in the shape of a cuboid. the hall is 20m long, 15m wide and 4m high. the community hall needs re-decorating with new paint for the walls and ceiling, and new tiles on the floor. a 20l tin of paint covers 40 squared metres and costs £15. 1 squared metre floor tiles cost £3 each. work out the total cost of paint and tiles needed to decorate the community hall.

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Answer:

the total cost of paint and tiles needed to decorate the community hall is £105 + £900 = £1005.

Step-by-step explanation:

Area of the walls:

- The community hall is shaped like a cuboid, so the walls are the four rectangular sides.

- The length and height of each wall are given as 20m and 4m, respectively.

- The total area of the walls is the sum of the areas of all four walls: 2 * (length * height + width * height).

- Substituting the given values, we have: 2 * (20m * 4m + 15m * 4m).

2. Area of the ceiling:

- The ceiling is also a rectangle, with the same length and width as the floor.

- The area of the ceiling is given by length * width: 20m * 15m.

3. Area of the floor:

- The area of the floor is the same as the area of the ceiling: 20m * 15m.

Now, let's calculate the areas:

1. Area of the walls:

- 2 * (20m * 4m + 15m * 4m) = 2 * (80m^2 + 60m^2) = 2 * 140m^2 = 280m^2.

2. Area of the ceiling:

- 20m * 15m = 300m^2.

3. Area of the floor:

- 20m * 15m = 300m^2.

Next, we can calculate the number of tins of paint and tiles needed and their costs:

- Number of tins of paint needed = Area of walls / Coverage per tin = 280m^2 / 40m^2/tin = 7 tins.

- Cost of paint = Number of tins * Cost per tin = 7 tins * £15/tin = £105.

- Number of tiles needed = Area of floor / Area per tile = 300m^2 / 1m^2/tile = 300 tiles.

- Cost of tiles = Number of tiles * Cost per tile = 300 tiles * £3/tile = £900.

Therefore, the total cost of paint and tiles needed to decorate the community hall is £105 + £900 = £1005.

yes, i am a student

I REALLY NEED HELP ON THIS, PLEASE!!!!

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1. $2d+3$

2.
$n\cdot 7\n7n$

What are the common factors of 60 36 and 24?

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1, 2, 3, 4, 6, 12 :)

I need urgent help with this. please write out the working and the answer thank you

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1.   C = ⁵/₉F - 32
9C = 9(⁵/₉F - 32)
  9C = 9(⁵/₉F) - 9(32)
  9C = 5F - 288
- 5F - 5F
  9C - 5F = -288
- 9C   - 9C
-5F = -9C - 288
-5 -5
F = 1⁴/₅C + 57.6

2. $H = w + (50)/(m^(2))$
   $m^(2)(H) = m^(2)((50)/(m^(2)))$
   $Hm^(2) = 50$
   $(Hm^(2))/(H) = (50)/(H)$
   $m^(2) = (50)/(H)$
   $\sqrt{m^(2)} = \sqrt{(50)/(H)}$
   $m = (√(50))/(√(H))$
   $m = (5√(2))/(√(H))$
   $m = (5√(2))/(√(H)) * (√(H))/(√(H))$
   $m = (5√(2H))/(H)$

3. A = 5 + 4√x
- 5 - 5
  A - 5 = 4√x
  (A - 5)² = (4√x)²
  A² - 10A + 25 = 16x
16 16
¹/₆A² - ⁵/₈A + 1⁹/₁₆ = x

4. $A = (b + c)/(b)$
   $Ab = b((b + c)/(b))$
   $Ab = b + c$
   $Ab - b = c$
   $b(A) - b(1) = c$
   $b(A - 1) = c$
   $(b(A - 1))/(A - 1) = (c)/(A - 1)$
   $b = (c)/(A - 1)$

3p4 ( 4p4 + 7p3 + 1 )

Answers

$3p^4 (4p^4 + 7p^3 + 1) \n \n 3p^4 * 4p^4 + 3p^4 * 7p^3 + 3p^4 \n \n 3 * 4p^4p^4 + 3p^4 * 7p^3 + 3p^4 \n \n 3 * 4p^(4 + 4) + 3p^4 * 7p^3 + 3p^4 \n \n 3 * 4p^8 + 3p^4 * 7p^3 + 3p^4 \n \n 12p^8 + 3p^4 * 7p^3 + 3p^4 \n \n 12p^8 + 3 * 7p^4p^4 + 3p^4 \n \n 12p^8 + 3 * 7p^(4 + 3) + 3p^4 \n \n 12p^8 + 3 * 7p^7 + 3p^4 \n \n 12p^8 + 21p^7 + 3p^4$

The answer is: 12p^8 + 21p^7 + 3p^4.

3p⁴(4p⁴ + 7p³ + 1)
3p⁴(4p⁴) + 3p⁴(7p⁴) + 3p⁴(1)
12p⁸ + 21p⁷ + 3p⁴

(x^2 - x^(1/2))/(1-x^(1/2))

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$\frac { \left( { x }^( 2 )-{ x }^{ \frac { 1 }{ 2 } } \right) }{ \left( 1-{ x }^{ \frac { 1 }{ 2 } } \right) }$

$\n \n =\frac { \left( { x }^( 2 )-\sqrt { x } \right) }{ \left( 1-\sqrt { x } \right) } \cdot 1$

$\n \n =\frac { \left( { x }^( 2 )-\sqrt { x } \right) }{ \left( 1-\sqrt { x } \right) } \cdot \frac { \left( 1+\sqrt { x } \right) }{ \left( 1+\sqrt { x } \right) }$

$\n \n =\frac { { x }^( 2 )+{ x }^( 2 )\sqrt { x } -\sqrt { x } -x }{ 1+\sqrt { x } -\sqrt { x } -x }$

$\n \n =\frac { -\sqrt { x } \left( 1-{ x }^( 2 ) \right) -x\left( 1-x \right) }{ \left( 1-x \right) }$

$\n \n =\frac { -\sqrt { x } \left( 1+x \right) \left( 1-x \right) -x\left( 1-x \right) }{ \left( 1-x \right) }$

$\n \n =\frac { \left( 1-x \right) \left\{ -\sqrt { x } \left( 1+x \right) -x \right\} }{ \left( 1-x \right) }$

$\n \n =-\sqrt { x } \left( 1+x \right) -x\n \n =-{ x }^{ \frac { 1 }{ 2 } }\left( 1+{ x }^{ \frac { 2 }{ 2 } } \right) -x$

$\n \n =-{ x }^{ \frac { 1 }{ 2 } }-{ x }^{ \frac { 3 }{ 2 } }-x\n \n =-\sqrt { x } -\sqrt { { x }^( 3 ) } -x$

x² - x^(1/2) = x²
1 - x^(1/2)

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