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4 years ago

Help thx.............. :)

Mathematics

2 answers:

photoshop1234 [79]4 years ago

7 0

For number 7 it’s basically 4x3 and 4x6 and for number 8 it’s 4x7 and 2x7
For number 9 it’s 5x3, 5x6, 5x2

Rainbow [258]4 years ago

4 0

^^^^^^^^^^^^^
What that guy said

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Two competitive neighbours build rectangular pools that cover the same area but are different shapes. Pool A has a width of (x +

GenaCL600 [577]

<u>Answer: </u>

a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

b) Area of pool A is equal to area of pool B equal to 24.44 meters.

<u> Solution: </u>

Let’s first calculate area of pool A .

Given that width of the pool A = (x+3)

Length of the pool A is 3 meter longer than its width.

So length of pool A = (x+3) + 3 =(x + 6)

Area of rectangle = length x width

So area of pool A =(x+6) (x+3) ------(1)

Let’s calculate area of pool B

Given that length of pool B is double of width of pool A.

So length of pool B = 2(x+3) =(2x + 6) m

Width of pool B is 4 meter shorter than its length,

So width of pool B = (2x +6 ) – 4 = 2x + 2

Area of rectangle = length x width

So area of pool B =(2x+6)(2x+2) ------(2)

Since area of pool A is equal to area of pool B, so from equation (1) and (2)

(x+6) (x+3) =(2x+6) (2x+2)

On solving above equation for x

(x+6) (x+3) =2(x+3) (2x+2)

x+6 = 4x + 4

x-4x = 4 – 6

x = $\frac{2}{3}$

Dimension of pool A

Length = x+6 = ( $\frac{2}{3}$ ) +6 = 6.667m

Width = x +3 = ( $\frac{2}{3}$ ) +3 = 3.667m

Dimension of pool B

Length = 2x +6 = 2( $\frac{2}{3}$ ) + 6 = $\frac{22}{3}$ = 7.333m

Width = 2x + 2 = 2( $\frac{2}{3}$ ) + 2 = $\frac{10}{3}$ = 3.333m

Verifying the area:

Area of pool A = ( $\frac{20}{3}$ ) x ( $\frac{11}{3}$ ) = $\frac{220}{9}$ = 24.44 meter

Area of pool B = ( $\frac{22}{3}$ ) x ( $\frac{10}{3}$ ) = $\frac{220}{9}$ = 24.44 meter

Summarizing the results:

(a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

(b)Area of pool A is equal to Area of pool B equal to 24.44 meters.

5 0

4 years ago

An open top box is to be built with a rectangular base whose length is twice its width and with a volume of 36 ft 3 . Find the d

denpristay [2]

Answer:

The dimensions of the box that minimize the materials used is $6\times 3\times 2\ ft$

Step-by-step explanation:

Given : An open top box is to be built with a rectangular base whose length is twice its width and with a volume of 36 ft³.

To find : The dimensions of the box that minimize the materials used ?

Solution :

An open top box is to be built with a rectangular base whose length is twice its width.

Here, width = w

Length = 2w

Height = h

The volume of the box V=36 ft³

i.e. $w\times 2w\times h=36$

$h=\frac{18}{w^2}$

The equation form when top is open,

$f(w)=2w^2+2wh+2(2w)h$

Substitute the value of h,

$f(w)=2w^2+2w(\frac{18}{w^2})+2(2w)(\frac{18}{w^2})$

$f(w)=2w^2+\frac{36}{w}+\frac{72}{w}$

$f(w)=2w^2+\frac{108}{w}$

Derivate w.r.t 'w',

$f'(w)=4w-\frac{108}{w^2}$

For critical point put it to zero,

$4w-\frac{108}{w^2}=0$

$4w=\frac{108}{w^2}$

$w^3=27$

$w^3=3^3$

$w=3$

Derivate the function again w.r.t 'w',

$f''(w)=4+\frac{216}{w^3}$

For w=3, $f''(3)=4+\frac{216}{3^3}=12 >0$

So, it is minimum at w=3.

Now, the dimensions of the box is

Width = 3 ft.

Length = 2(3)= 6 ft

Height = $\frac{18}{3^2}=2\ ft$

Therefore, the dimensions of the box that minimize the materials used is $6\times 3\times 2\ ft$

4 0

4 years ago

Which is greater thirty six and nine thousandths or four tens

malfutka [58]

Answer:

four tens

Step-by-step explanation:

four tens is greater because:

four tens = 40

thirty six and nine thousandths = 36.009

36.009 < 40

6 0

4 years ago

At Mountain Park Middle School, there are 250 seventh-grade students. Currently, 32% of those students are in

sergij07 [2.7K]

A =250 - (250 x 0.32)

3 0

3 years ago

Help plz i need help with this please provide an explanation

miss Akunina [59]

It’s easier to see and solve if you draw the picture
To find LD you need to subtract CD to CL
CD - CL = LD
14.2 - 5.3 = 8.9
=> LD = 8.9

5 0

3 years ago