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Lubov Fominskaja [6]
1 year ago
5

Taylor uses tape to mark a square play area in the basement for her daughter. The area measures 28 ft2. Is the side length of th

e square rational or irrational? Explain.
Mathematics
2 answers:
jeka941 year ago
5 0

Answer:

28 is your area and it is a rational number it's never ending number

Step-by-step explanation:

joja [24]1 year ago
4 0

The side length of the square is a rational.

What is the area of the basement?

Any lower storey of a structure that is partially or completely below the lowest contiguous proposed ground level is referred to as a basement floor, provided that the portion of the storey above ground level does not exceed 75 cm.

Given: Taylor uses tape to mark a square play area in the basement for her daughter.

The area measures 28 ft2.

We know that the area of square is side length square.

As the area measures 28 square feet, so the side length of the square is, 28 feet.

Since 28 is a rational number.

Therefore, the side length of the square is a rational.

To know more about the area of basement, click on the link

brainly.com/question/26791387

#SPJ2

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MATH HELP!!! 100PTS!!!
Lostsunrise [7]

Answer:

x =\dfrac{20 +\sqrt{454}}{18}, \quad \dfrac{20 -\sqrt{454}}{18}

Step-by-step explanation:

<u>Given functions</u>:

  p(x)=2x-4

  r(x)=\dfrac{6x-1}{9x+1}

Solve for p(x) = r(x):

\begin{aligned}p(x) & = r(x)\\\implies 2x-4 & = \dfrac{6x-1}{9x+1}\\(2x-4)(9x+1)&=6x-1\\18x^2+2x-36x-4&=6x-1\\18x^2-40x-3&=0\end{aligned}

As the found quadratic equation cannot be factored, use the Quadratic Formula to solve for x:

<u>Quadratic Formula</u>

x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0

Therefore:

a=18, \quad b=-40, \quad c=-3

Substitute the values of a, b and c into the <u>quadratic formula</u> and solve for x:

\begin{aligned}\implies x & =\dfrac{-(-40) \pm \sqrt{(-40)^2-4(18)(-3)} }{2(18)}\\\\& =\dfrac{40 \pm \sqrt{1816}}{36}\\\\& =\dfrac{40 \pm \sqrt{4 \cdot 454}}{36}\\\\& =\dfrac{40 \pm \sqrt{4}\sqrt{454}}{36}\\\\& =\dfrac{40 \pm2\sqrt{454}}{36}\\\\& =\dfrac{20 \pm\sqrt{454}}{18}\end{aligned}

Therefore, the solutions are:

x =\dfrac{20 +\sqrt{454}}{18}, \quad \dfrac{20 -\sqrt{454}}{18}

Learn more about quadratic equations here:

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The number of bacteria growing in an incubation culture increases with time according to n left parenthesis t right parenthesis
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