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inna [77]
4 years ago
13

Please help me!! I don't get this question! I will mark Brainliest!

Mathematics
2 answers:
TEA [102]4 years ago
5 0

Answer:

Part A:

Formula for circumference: C = diameter x pi

=> pi = C/diameter

=> pi(A) = 12.56/4 =3.14

=> pi(B) = 6.28/2 = 3.14

Part B:

Formula for area: A = radius^2 x pi = (diameter/2)^2 x pi

=> pi = A/(diameter/2)^2

=> pi(A) = 12.56/(4/2)^2 = 12.56/4 = 3.14

=> pi(B) = 3.14/(2/2)^2 = 3.14/1 = 3.14

Part C:

pi(A) = pi(B) = 3.14 in both ways of calculation derived from circumstance and area.

=> pi is a constant value: 3.14

bagirrra123 [75]4 years ago
3 0

Answer:

A) pi = 3.14

B) pi = 3.14

C) pi has the same value for all calculations, pi is a constant

Step-by-step explanation:

Circumference = pi × d

12.56 = 4 × pi

pi = 3.14

6.28 = 2 × pi

pi = 3.14

Area = pi × r²

12.56 = pi × (4/2)²

12.56 = 4pi

pi = 3.14

3.14 = pi × (2/2)²

3.14 = pi

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The Least Common Multiple ( LCM ) is also referred to as the Lowest Common Multiple (LCM) and Least Common Divisor (LCD). For two integers a and b, denoted LCM (a,b), the LCM is the smallest positive integer that is evenly divisible by both a and b. For example, LCM (2,3) = 6 and LCM (6,10) = 30.

Step-by-step explanation:

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$ 3.34

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Likurg_2 [28]

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Step-by-step explanation:

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Solving Rational equations. LCD method. Show work. Image attached.
posledela

(k-2)(k-6)=k^2-2k-6k+12=k^2-8k+12

So in order to get all the fractions to have a common denominator, we need to multiply \dfrac k{k-2} by \dfrac{k-6}{k-6}, and \dfrac1{k-6} by \dfrac{k-2}{k-2}:

\dfrac k{k-2}\cdot\dfrac{k-6}{k-6}=\dfrac{k(k-6)}{(k-2)(k-6)}=\dfrac{k^2-6k}{k^2-8k+12}

\dfrac1{k-6}\cdot\dfrac{k-2}{k-2}=\dfrac{k-2}{(k-2)(k-6)}=\dfrac{k-2}{k^2-8k+12}

Now,

\dfrac4{k^2-8k+12}=\dfrac{(k^2-6k)+(k-2)}{k^2-8k+12}

As long as k\neq2 and k\neq6 (which we can't have because otherwise k^2-8k+12=0), we can cancel k^2-8k+12 in the denominators on both sides:

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We can factorize the right side:

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which tells us that k=6 and k=-1 are solutions.

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4 years ago
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