In the picture given below, can see a rectangular in ground swimming pool installed by a family in their backyard. there is a concrete sidewalk around the pool of with X metre. the outside edges of the side of measure 7 m and 12 m. the area of the pool is 36 square metre​

The quadratic equation in terms of x is 4x²-38x+48 = 0 and the width of the sidewalk is 1.5 m.

Given,

The outer edges of the swimming pool are 7m and 12 m.

The width of the sidewalk is x m.

The area of the pool = 36 sq. m

To Find,

The quadratic equation in terms of x, and the width of the sidewalk.

Solution,

Since the width of the sidewalk is x m.

So the dimensions of the pool will be (7-2x) and (12-2x).

Now,

Area of the pool = 36 sq. m

(7-2x)(12-2x) = 36

84-14x-24x+4x² = 36

4x²-38x+48 = 0

4(x-8)(x-3/2) = 0

x = 8, 1.5

The value of x can't be 8 as it will make the dimensions of the pool negative

So, the value of x is 1.5 m.

Hence, the quadratic equation in terms of x is 4x²-38x+48 = 0 and the width of the sidewalk is 1.5 m.

Answer:

The quadratic equation in 'x' is 2x² -19x +24 = 0

Width of the sidewalk = 1.5m

Explanation:

Given,

Area of the pool = 36sq.m.

width of the sidewalk = 'x'm

Length of the pool including the sidewalk = 7m

Breadth of the pool including the sidewalk = 12m

Required to find,

Quadratic equation in terms of 'x'

Width of the sidewalk

Formula to be used

Area of the rectangle = length x breadth

Length of the pool = 7 -2x

Breadth of the pool = 12 -2x

Area of the pool = (7-2x)(12-2x)

(7-2x)(12-2x) = 36

84 - 14x -24x +4x² = 36

4x² - 38x + 48 = 0

2x² - 19x + 24 = 0

Hence,

The quadratic equation in 'x' is  2x² - 19x + 24 = 0

To find 'x'

To solve the quadratic equation by splitting the middle term, we need to find two numbers such that their sum = -19 and product = 48

Two such numbers are = -16 and -3

2x (x-8) - 3(x-8) = 0

(2x-3)(x-8) = 0

2x -3 = 0 or x = 8

x =[tex]\frac{3}{2}[/tex]  = 1.5 or x = 8

x = 8, is not possible because the length of the pool including the sidewalk is 7m.

So x = 1.5m

∴ Width of the sidewalk = 1.5m

Answer:

12, 18 , ......78

take an =

get n = 12

then find S12 to get it as 540

Answer: Sum of first 12 2 digit numbers which are multiple of 6 is 540.

Step-by-step explanation:

Given series : 12, 18, 24, 30, 36, .....

This is an Arithmetic Progression.

[tex]a_{1} = 12[/tex]

common difference = 6

∴ [tex]S_{12} = \frac{n}{2} (2*a_{1} + (n - 1) * d)[/tex]

         = [tex]\frac{12}{2} (2*12 + (12 - 1) * 6)[/tex]

         = 540

Answer:

(a)

[tex]2 {x}^{2} - 19x + 24 = 0[/tex]

(b)

[tex]x = \frac{3}{2} [/tex]

Step-by-step explanation:

Refer to the attachment↑↑

Answer: dimensions of pool are 4m and 9m.

Step-by-step explanation:

• Given,the width of the path is x m.

• And the outside edges of pool ,easures 7m and 12m.

• So,the dimensions of pool will decrease by 2x i.e., x from each side.

• Dimensions of pool= (7-2x)m and (12-2x)m

• Given,area of pool= 36 sq.m

• Now,according to question

       (7-2x)×(12-2x)=36

    ⇒7(12-2x)-2x(12-2x)=36

    ⇒84-14x-24x+4=36

    ⇒4-38x+84-36=0

    ⇒4-38x+48=0

    ⇒2(2-19x+24)=0

    ⇒(2-19x+24)=0

    ⇒2-16x-3x+24=0

    ⇒2x(x-8)-3(x-8)=0

    ⇒(2x-3)(x-8)=0

    ⇒If 2x-3=0 then x=  and if x-8=0 then x=8(not possible)

     ∴ x=1.5m

• (7-2x)=(7-2*1.5)=(7-3)=4m and (12-2x)=(12-3)=9m

• So, the dimensions of pool are (7-2x)m and (12-2x)m i.e.,4m and 9m.

Answer:Is it correct

Step-by-step explanation:

Step-by-step explanation:

Median = l + [ (n/2 - c)/f ] × h

here, l = lower class

         n = total number of observation

         c = cumulative frequency

          f = frequency

         h = class size

 Class interval       Frequency        Cumulative Frequency

      0-15                         17                                17

      15-30                      35                               52

      30-45                      40                              92

      45-60                      18                               110

       60-75                      m                               110+m

       75-90                       2                               112+m

    33 = 30 + [ (112 + m)/2 - 52 ]/40 × 15

    3 =  (112 + m - 104)/80 × 15

    1/5 = (8 + m)/80

     16 = 8 + m

     m = 8