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Nataliya [291]
3 years ago
10

the groundskeeper at Orangefield must mow a circular field the diameter of the field is 250 yards what is the area of the field

that the groundskeeper must mow.
Mathematics
1 answer:
andrey2020 [161]3 years ago
4 0
The answer is approximately 49087.385212340519350978802863742.

A= pi times the radius squared. r=125
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Hi! Algebra two question, rational exponents.
ivanzaharov [21]

Simplifying \left(x^2+8xy+16y^2\right)^{\frac{1}{3}}.\left(x+4y\right)^{\frac{1}{3}} we get \left(x+4y\right)}

Step-by-step explanation:

We need to simplify:

\left(x^2+8xy+16y^2\right)^{\frac{1}{3}}.\left(x+4y\right)^{\frac{1}{3}}

For solving, we will first simplify the terms inside the exponent and then solve exponent

First Solving: \left(x^2+8xy+16y^2\right)}.\left(x+4y\right)}

Using the formula: (a^2+2ab+b^2)=(a+b)^2

\left(x^2+8xy+16y^2\right)}.\left(x+4y\right)}\\\left((x)^2+2(x)(4y)+(4y)^2\right)}.\left(x+4y\right)}\\\left(x+4y\right)^2}.\left(x+4y\right)}

Using exponent rule:

a^m.a^n=a^{m+n}

\left(x+4y\right)^{2+1}}\\\left(x+4y\right)^3}

Now,

(\left(x+4y\right)^3})^{\frac{1}{3}}\\Simplifiying:\\\left(x+4y\right)}^{\frac{3}{3}}\\\left(x+4y\right)}

So, simplifying \left(x^2+8xy+16y^2\right)^{\frac{1}{3}}.\left(x+4y\right)^{\frac{1}{3}} we get \left(x+4y\right)}

Keywords: Solving Rational exponents:

Learn more about Solving Rational exponents at:

  • brainly.com/question/13174254
  • brainly.com/question/13174255
  • brainly.com/question/5639299
  • brainly.com/question/729447

#learnwithBrainly

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The profit earned by a hot dog stand is a linear function of the number of hot dogs sold. It costs the owner $48 dollars each
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Answer:

y=2x-48

Step-by-step explanation:

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So basically its all the numbers that are can be an answer for example if ur looking at a dice you set of all possible outcomes will be {1,2,3,4,5,6} because a dice has 6 possible out comes
3 0
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Thank you all -Lily<br> But i still have another math test to do
ladessa [460]
Goodluck on your math test
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Answer:

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Step-by-step explanation:

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