Question

PLS ANSWER IN THIS MATH

Answer 1

Question 1.

The dimensions of the rectangle is given as 3x-5 by 2x+8.

The perimeter of a rectangle is given by:

`P= 2(l + w)`

We substitute the dimensions to get:

`P= 2(3x - 5 + 2x + 8)`

We simplify to obtain:

`P= 2(5x + 3)`

`P= 10x + 6`

If each foot costs $5 then x feet will cost 5x dollars.

Therefore the total cost of fencing will be;

`10(5x) + 6 * 5= 50x + 30`

The total cost in terms of x is 50x+30

Question 2.

The area of the rectangle is given as

`{x}^(2) - 2x - 35`

We can factor to find the possible dimensions of this rectangle.

We split the middle term to get:

`{x}^(2) + 5x - 7x - 35`

We factor by grouping:

`x(x +5 ) - 7(x + 5)`

We factor further to get:

`(x - 7)(x + 5)`

The possible dimensions are x-7 by x+5

Answer 2

Question # 1

Answer:

It would cost
`50x+30` dollars to put fencing around this garden.

Explanation:

In Linda's garden each side is:

• either
  `3x-5` feet or
  `2x+8` feet

so

We need to determine the perimeter of the rectangle in order to figure out how much fencing there is.

`P=\left(3x-5\right)+\left(2x+8\right)+\left(3x-5\right)+\left(2x+8\right)`

As price per foot of fencing is 5 dollars.

So multiply this by 5 dollars.

`5\left(P\right)=5\left[\left(3x-5\right)+\left(2x+8\right)+\left(3x-5\right)+\left(2x+8\right)\right]`

solving

`5\left[\left(3x-5\right)+\left(2x+8\right)+\left(3x-5\right)+\left(2x+8\right)\right]`

`=5\left(\left(3x-5\right)+\left(2x+8\right)+\left(3x-5\right)+\left(2x+8\right)\right)`

`\mathrm{Remove\:parentheses}:\quad \left(a\right)=a`

`=5\left(3x-5+2x+8+3x-5+2x+8\right)...[A]`

`\mathrm{Simplify}\:3x-5+2x+8+3x-5+2x+8`

`3x-5+2x+8+3x-5+2x+8`

`=3x+2x+3x+2x-5+8-5+8`

`=10x-5+8-5+8`

`=10x+6`

so Equation [A] becomes

`=5\left(10x+6\right)`

`=50x+30`

Therefore, it would cost
`50x+30` dollars to put fencing around this garden.

Question # 2

Answer:

The possible dimensions of the rectangle are
`\left(x-5\right)\left(x+7\right)`.

Explanation:

As the rectangle has the area =
`x^2+2x-35`

As

Area = Length × Width

The dimensions of the rectangle could be found by factoring the area:

`x^2+2x-35`

`\mathrm{Break\:the\:expression\:into\:groups}`

`=\left(x^2-5x\right)+\left(7x-35\right)`

`\mathrm{Factor\:out\:}x\mathrm{\:from\:}x^2-5x\mathrm{:\quad }x\left(x-5\right)`

`\mathrm{Factor\:out\:}7\mathrm{\:from\:}7x-35\mathrm{:\quad }7\left(x-5\right)`

`=x\left(x-5\right)+7\left(x-5\right)`

`\mathrm{Factor\:out\:common\:term\:}x-5`

`=\left(x-5\right)\left(x+7\right)`

Therefore, the possible dimensions of the rectangle are
`\left(x-5\right)\left(x+7\right)`.